比赛链接：传送门

前期大顺风，2：30金区中游。后期开题乏力，掉到银尾。4：59绝杀I，但罚时太高卡在银首。

Problem A - Dogs and Cages 00：09：45 （+） Solved by Dancepted

算了半天发现就是n-1，被队友喷死，差点气哭。

Problem E - Evil Forest 00：16：54 （+） Solved by xk

xk大喊签到，就过了。

代码：

  1. #include <iostream>
  2. #include <cmath>
  3. #include <map>
  4. #include <algorithm>
  5. #include <cstdio>
  6. #include <cstring>
  7. #include <set>
  8. #include <vector>
  9. #include <string>
  10. #include <queue>
  11. #include <stack>
  12. #include <iomanip>
  13. #define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
  14. #define N 100005
  15. #define M 100005
  16. #define INF 0x3f3f3f3f
  17. #define mk(x) (1<<x) // be conscious if mask x exceeds int
  18. #define sz(x) ((int)x.size())
  19. #define upperdiv(a,b) (a/b + (a%b>0))
  20. #define mp(a,b) make_pair(a, b)
  21. #define endl '\n'
  22. #define lowbit(x) (x&-x)
  23.  
  24. using namespace std;
  25. typedef long long ll;
  26. typedef double db;
  27.  
  28. /** fast read **/
  29. template <typename T>
  30. inline void read(T &x) {
  31.     x = ; T fg = ; char ch = getchar();
  32.     while (!isdigit(ch)) {
  33.         if (ch == '-') fg = -;
  34.         ch = getchar();
  35.     }
  36.     while (isdigit(ch)) x = x*+ch-'', ch = getchar();
  37.     x = fg * x;
  38. }
  39. template <typename T, typename... Args>
  40. inline void read(T &x, Args &... args) { read(x), read(args...); }
  41. template <typename T>
  42. inline void write(T x) {
  43.     int len = ; char c[]; if (x < ) putchar('-'), x = -x;
  44.     do{++len; c[len] = x% + '';} while (x /= );
  45.     for (int i = len; i >= ; i--) putchar(c[i]);
  46. }
  47. template <typename T, typename... Args>
  48. inline void write(T x, Args ... args) { write(x), write(args...); }
  49.  
  50. int main() {
  51.     fast;
  52.     int T, kase = ;
  53.     cin >> T;
  54.     while (T--) {
  55.         int n;
  56.         cin >> n;
  57.         int ans = ;
  58.         for(int i = ; i < n; i++)
  59.         {
  60.             int x;
  61.             cin >> x;
  62.             ans += x + upperdiv(x, );
  63.         }
  64.         cout << "Case #" << (kase++) << ": " << ans << endl;
  65.     }
  66.     return ;
  67. }

Problem C - Rich Game 00：32：42 （+） Solved by xk

xk又喊了一声签到！

代码：

  1. #include <iostream>
  2. #include <cmath>
  3. #include <map>
  4. #include <algorithm>
  5. #include <cstdio>
  6. #include <cstring>
  7. #include <set>
  8. #include <vector>
  9. #include <string>
  10. #include <queue>
  11. #include <stack>
  12. #include <iomanip>
  13. #define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
  14. #define N 100005
  15. #define M 100005
  16. #define INF 0x3f3f3f3f
  17. #define mk(x) (1<<x) // be conscious if mask x exceeds int
  18. #define sz(x) ((int)x.size())
  19. #define upperdiv(a,b) (a/b + (a%b>0))
  20. #define mp(a,b) make_pair(a, b)
  21. #define endl '\n'
  22. #define lowbit(x) (x&-x)
  23.  
  24. using namespace std;
  25. typedef long long ll;
  26. typedef double db;
  27.  
  28. /** fast read **/
  29. template <typename T>
  30. inline void read(T &x) {
  31.     x = ; T fg = ; char ch = getchar();
  32.     while (!isdigit(ch)) {
  33.         if (ch == '-') fg = -;
  34.         ch = getchar();
  35.     }
  36.     while (isdigit(ch)) x = x*+ch-'', ch = getchar();
  37.     x = fg * x;
  38. }
  39. template <typename T, typename... Args>
  40. inline void read(T &x, Args &... args) { read(x), read(args...); }
  41. template <typename T>
  42. inline void write(T x) {
  43.     int len = ; char c[]; if (x < ) putchar('-'), x = -x;
  44.     do{++len; c[len] = x% + '';} while (x /= );
  45.     for (int i = len; i >= ; i--) putchar(c[i]);
  46. }
  47. template <typename T, typename... Args>
  48. inline void write(T x, Args ... args) { write(x), write(args...); }
  49.  
  50. int main() {
  51.     fast;
  52.     int T, kase = ;
  53.     cin >> T;
  54.     while (T--) {
  55.         int x, y, k;
  56.         cin >> x >> y >> k;
  57.         int money = , ans = ;
  58.         if(x > y) ans = k;
  59.         else
  60.         {
  61.             for(int i = ; i < k; i++)
  62.             {
  63.                 if(money +  * x >=  * y) {
  64.                     money +=  * x -  * y;
  65.                     ans++;
  66.                 }
  67.                 else {
  68.                     money +=  * x;
  69.                 }
  70.             }
  71.         }
  72.         cout << "Case #" << (kase++) << ": " << ans << endl;
  73.     }
  74.     return ;
  75. }

Problem K - Knightmare 00：46：33 （+） Solved by Dancepted

步数较少的时候可以往回走，把没走过的地方填上，步数增长到一定程度时，答案差不多会均匀增长。

我猜测是小数据打表，大数据是个公式之类的。

开完A趁电脑没人就打了个表，发现和我的猜测是一致的，果断交了一发就过了。

代码：

  1. #include <iostream>
  2. #include <cmath>
  3. #include <map>
  4. #include <algorithm>
  5. #include <cstdio>
  6. #include <cstring>
  7. #include <set>
  8. #include <stdio.h>
  9. #include <utility>
  10. #include <vector>
  11. #include <string>
  12. #include <queue>
  13. #include <stack>
  14. #include <iomanip>
  15. #define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
  16. #define N 1005
  17. #define M 100005
  18. #define INF 0x3f3f3f3f
  19. #define mk(x) (1<<x) // be conscious if mask x exceeds int
  20. #define sz(x) ((int)x.size())
  21. #define upperdiv(a,b) (a/b + (a%b>0))
  22. #define mp(a,b) make_pair(a, b)
  23. #define endl '\n'
  24. #define lowbit(x) (x&-x)
  25.  
  26. using namespace std;
  27. typedef __int128 ll;
  28. typedef double db;
  29.  
  30. /** fast read **/
  31. template <typename T>
  32. inline void read(T &x) {
  33.     x = ; T fg = ; char ch = getchar();
  34.     while (!isdigit(ch)) {
  35.         if (ch == '-') fg = -;
  36.         ch = getchar();
  37.     }
  38.     while (isdigit(ch)) x = x*+ch-'', ch = getchar();
  39.     x = fg * x;
  40. }
  41. template <typename T, typename... Args>
  42. inline void read(T &x, Args &... args) { read(x), read(args...); }
  43. template <typename T>
  44. inline void write(T x) {
  45.     int len = ; char c[]; if (x < ) putchar('-'), x = -x;
  46.     do{++len; c[len] = x% + '';} while (x /= );
  47.     for (int i = len; i >= ; i--) putchar(c[i]);
  48. }
  49. template <typename T, typename... Args>
  50. inline void write(T x, Args ... args) { write(x), write(args...); }
  51.  
  52. const int ans[] = {, , , , , };
  53. int main() {
  54.     int T; cin >> T;
  55.     int kase = ;
  56.     while (T--) {
  57.         ll n; read(n);
  58.         printf("Case #%d: ", kase++);
  59.         if (n <= ) {
  60.             write(ans[n]);
  61.         }
  62.         else {
  63.             ll res =  * n * n -  * n + ;
  64.             write(res);
  65.         }
  66.         putchar('\n');
  67.     }
  68.  
  69.     return ;
  70. }

Problem J - Subway Chasing 01：42：00 （+） Solved by xk & lh

好像是差分约束什么的。图论这种事相信队友就完了。

代码：

  1. #include <iostream>
  2. #include <cmath>
  3. #include <map>
  4. #include <algorithm>
  5. #include <cstdio>
  6. #include <cstring>
  7. #include <set>
  8. #include <vector>
  9. #include <string>
  10. #include <queue>
  11. #include <stack>
  12. #include <iomanip>
  13. #define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
  14. #define N 100005
  15. #define M 100005
  16. #define INF 0x3f3f3f3f
  17. #define mk(x) (1<<x) // be conscious if mask x exceeds int
  18. #define sz(x) ((int)x.size())
  19. #define upperdiv(a,b) (a/b + (a%b>0))
  20. #define mp(a,b) make_pair(a, b)
  21. #define endl '\n'
  22. #define lowbit(x) (x&-x)
  23.  
  24. using namespace std;
  25. typedef long long ll;
  26. typedef double db;
  27.  
  28. /** fast read **/
  29. template <typename T>
  30. inline void read(T &x) {
  31.     x = ; T fg = ; char ch = getchar();
  32.     while (!isdigit(ch)) {
  33.         if (ch == '-') fg = -;
  34.         ch = getchar();
  35.     }
  36.     while (isdigit(ch)) x = x*+ch-'', ch = getchar();
  37.     x = fg * x;
  38. }
  39. template <typename T, typename... Args>
  40. inline void read(T &x, Args &... args) { read(x), read(args...); }
  41. template <typename T>
  42. inline void write(T x) {
  43.     int len = ; char c[]; if (x < ) putchar('-'), x = -x;
  44.     do{++len; c[len] = x% + '';} while (x /= );
  45.     for (int i = len; i >= ; i--) putchar(c[i]);
  46. }
  47. template <typename T, typename... Args>
  48. inline void write(T x, Args ... args) { write(x), write(args...); }
  49.  
  50. const int maxn =  + ;
  51.  
  52. struct edge
  53. {
  54.     int v, l;
  55. };
  56.  
  57. ll dis[maxn];
  58. int cnt[maxn];
  59. bool inq[maxn];
  60. vector<edge> g[maxn];
  61.  
  62. void addedge(int u, int v, int l)
  63. {
  64.     g[u].push_back(edge{v, l});
  65.     // g[v].push_back(edge{u, l});
  66. }
  67.  
  68. int n, m, x;
  69.  
  70. bool spfa()
  71. {
  72.     queue<int> q;
  73.     dis[] = ;
  74.     q.push();
  75.     cnt[] = inq[] = ;
  76.     while(!q.empty())
  77.     {
  78.         int u = q.front();
  79.         inq[u] = ;
  80.         q.pop();
  81.         for(auto &e : g[u])
  82.         {
  83.             if(dis[e.v] > dis[u] + e.l)
  84.             {
  85.                 dis[e.v] = dis[u] + e.l;
  86.  
  87.                 if(!inq[e.v]) {
  88.                     cnt[e.v]++;
  89.                     if(cnt[e.v] >= n) return false;
  90.                     q.push(e.v);
  91.                     inq[e.v] = ;
  92.                 }
  93.             }
  94.         }
  95.     }
  96.     return true;
  97. }
  98.  
  99. int main()
  100. {
  101.     fast;
  102.     int T, kase = ;
  103.     cin >> T;
  104.     while (T--)
  105.     {
  106.         cin >> n >> m >> x;
  107.         for(int i = ; i <= n; i++)
  108.         {
  109.             g[i].clear();
  110.             dis[i] = 1e18;
  111.             cnt[i] = inq[i] = ;
  112.         }
  113.         for(int i = ; i < n; i++)
  114.         {
  115.             addedge(i+, i, -);
  116.             addedge(i, i+, 2e9);
  117.         }
  118.         for(int i = ; i < m; i++)
  119.         {
  120.             int a, b, c, d;
  121.             cin >> a >> b >> c >> d;
  122.             if(a == b)
  123.             {
  124.                 if(c == d)
  125.                 {
  126.                     addedge(a, c, x);
  127.                     addedge(c, a, -x);
  128.                 }
  129.                 else
  130.                 {
  131.                     addedge(a, c, x - );
  132.                     addedge(d, a, -x - );
  133.                 }
  134.             }
  135.             else
  136.             {
  137.                 if(c == d)
  138.                 {
  139.                     addedge(b, c, x - );
  140.                     addedge(c, a, -x - );
  141.                 }
  142.                 else
  143.                 {
  144.                     addedge(b, c, x - );
  145.                     addedge(d, a, -x - );
  146.                 }
  147.             }
  148.         }
  149.         cout << "Case #" << (kase++);
  150.         if(!spfa()) cout << " IMPOSSIBLE\n";
  151.         else {
  152.             for(int i = ; i <= n; i++)
  153.                 cout << ' ' << dis[i] - dis[i - ];
  154.             cout << endl;
  155.         }
  156.     }
  157.     return ;
  158. }

Problem G - Alice’s Stamps 02：24：14 （-2） Solved by Dancepted （dp）

上机的时候思路有乱，调出来之后交上去因为N开大了MLE返回CE，贡献了两发罚时。

dfs枚举当前搜索过的最右端点和剩下的k的数量，记忆化一下，状态数最多就$O(n^{2})$。

HDU的T组数据比较玄学，差点没敢交。

代码：$O(n^{2})$

  1. #include <iostream>
  2. #include <cmath>
  3. #include <map>
  4. #include <algorithm>
  5. #include <cstdio>
  6. #include <cstring>
  7. #include <set>
  8. #include <vector>
  9. #include <string>
  10. #include <queue>
  11. #include <stack>
  12. #include <iomanip>
  13. #define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
  14. #define N 2005
  15. #define M 100005
  16. #define INF 0x3f3f3f3f
  17. #define mk(x) (1<<x)
  18. #define sz(x) ((int)x.size())
  19. #define upperdiv(a,b) (a/b + (a%b>0))
  20. #define mp(a,b) make_pair(a, b)
  21. #define endl '\n'
  22. #define lowbit(x) (x&-x)
  23.  
  24. using namespace std;
  25. typedef long long ll;
  26. typedef double db;
  27.  
  28. /** fast read **/
  29. template <typename T>
  30. inline void read(T &x) {
  31.     x = ; T fg = ; char ch = getchar();
  32.     while (!isdigit(ch)) {
  33.         if (ch == '-') fg = -;
  34.         ch = getchar();
  35.     }
  36.     while (isdigit(ch)) x = x*+ch-'', ch = getchar();
  37.     x = fg * x;
  38. }
  39. template <typename T, typename... Args>
  40. inline void read(T &x, Args &... args) { read(x), read(args...); }
  41. template <typename T>
  42. inline void write(T x) {
  43.     int len = ; char c[]; if (x < ) putchar('-'), x = -x;
  44.     do{++len; c[len] = x% + '';} while (x /= );
  45.     for (int i = len; i >= ; i--) putchar(c[i]);
  46. }
  47. template <typename T, typename... Args>
  48. inline void write(T x, Args ... args) { write(x), write(args...); }
  49.  
  50. struct Node{
  51.     int l, r;
  52.     bool operator < (const Node& x) const {
  53.         if (r == x.r) {
  54.             return l < x.l;
  55.         }
  56.         return r < x.r;
  57.     }
  58. };
  59.  
  60. int n, m, k;
  61. vector<Node> ns, ns1;
  62. int ans = ;
  63. int f[N][N];
  64. int dfs(int id, int r, int cnt, int resk) {
  65.     if (f[r][resk]) {
  66.         return f[r][resk];
  67.     }
  68.     if (resk == ) {
  69.         return ;
  70.     }
  71.     if (id >= sz(ns)) {
  72.         return ;
  73.     }
  74.     int tmp = dfs(id+, r, cnt, resk);
  75.     tmp = max(tmp, ns[id].r - max(ns[id].l-, r) + dfs(id+, ns[id].r, cnt + ns[id].r - max(ns[id].l-, r), resk-));
  76.     ans = max(ans, cnt + tmp);
  77.     return f[r][resk] = tmp;
  78. }
  79. bool ok[N];
  80. int l[N], r[N];
  81. int main() {
  82.     int T, kase = ;
  83.     cin >> T;
  84.     while (T--) {
  85.         ns.clear(), ns1.clear();
  86.         read(n, m, k);
  87.         for (int i = ; i <= m; i++) {
  88.             read(l[i], r[i]);
  89.             ok[i] = true;
  90.         }
  91.         for (int i = ; i <= m; i++) {
  92.             for (int j = ; j <= m; j++) if (i != j) {
  93.                 if (l[i] < l[j] && r[j] <= r[i]) {
  94.                     ok[j] = false;
  95.                 }
  96.             }
  97.         }
  98.         for (int i = ; i <= m; i++) if (ok[i]) {
  99.             ns.push_back(Node{l[i], r[i]});
  100.         }
  101.         sort(ns.begin(), ns.end());
  102.  
  103.         // init();
  104.         ans = ;
  105.         for (int i = ; i <= n; i++) {
  106.             for (int j = ; j <= k; j++) {
  107.                 f[i][j] = ;
  108.             }
  109.         }
  110.         dfs(, , , k);
  111.         printf("Case #%d: %d\n", kase++, ans);
  112.     }
  113.     return ;
  114. }

题解的做法是一个很好写的dp：

$f_{i, j}$表示最右边的覆盖点是i的情况下，用了j个集合的最大覆盖长度。

代码：$O(n^{2})$

  1. #include <iostream>
  2. #include <cmath>
  3. #include <map>
  4. #include <algorithm>
  5. #include <cstdio>
  6. #include <cstring>
  7. #include <set>
  8. #include <vector>
  9. #include <string>
  10. #include <queue>
  11. #include <stack>
  12. #include <iomanip>
  13. #define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
  14. #define N 2005
  15. #define M 100005
  16. #define INF 0x3f3f3f3f
  17. #define mk(x) (1<<x) // be conscious if mask x exceeds int
  18. #define sz(x) ((int)x.size())
  19. #define upperdiv(a,b) (a/b + (a%b>0))
  20. #define mp(a,b) make_pair(a, b)
  21. #define endl '\n'
  22. #define lowbit(x) (x&-x)
  23.  
  24. using namespace std;
  25. typedef long long ll;
  26. typedef double db;
  27.  
  28. /** fast read **/
  29. template <typename T>
  30. inline void read(T &x) {
  31.     x = ; T fg = ; char ch = getchar();
  32.     while (!isdigit(ch)) {
  33.         if (ch == '-') fg = -;
  34.         ch = getchar();
  35.     }
  36.     while (isdigit(ch)) x = x*+ch-'', ch = getchar();
  37.     x = fg * x;
  38. }
  39. template <typename T, typename... Args>
  40. inline void read(T &x, Args &... args) { read(x), read(args...); }
  41. template <typename T>
  42. inline void write(T x) {
  43.     int len = ; char c[]; if (x < ) putchar('-'), x = -x;
  44.     do{++len; c[len] = x% + '';} while (x /= );
  45.     for (int i = len; i >= ; i--) putchar(c[i]);
  46. }
  47. template <typename T, typename... Args>
  48. inline void write(T x, Args ... args) { write(x), write(args...); }
  49.  
  50. int mxr[N];
  51. int f[N][N];
  52. int main() {
  53.     int T; cin >> T;
  54.     for (int kase = ; kase <= T; kase++) {
  55.         int n, m, k; read(n, m, k);
  56.         memset(mxr, , (n+) * sizeof(int));
  57.         for (int i = ; i <= n; i++) {
  58.             for (int j = ; j <= k; j++) {
  59.                 f[i][j] = ;
  60.             }
  61.         }
  62.         for (int i = ; i <= m; i++) {
  63.             int l, r; read(l, r);
  64.             mxr[l] = max(mxr[l], r);
  65.         }
  66.         int ans = , r = ;
  67.         for (int i = ; i <= n; i++) {
  68.             r = max(mxr[i], r);
  69.             for (int j = ; j <= k; j++) {
  70.                 f[i][j] = max(f[i][j], f[i-][j]);
  71.                 ans = max(ans, f[i][j]);
  72.                 if (j+ <= k)
  73.                     f[r][j+] = max(f[r][j+], f[i-][j] + r - i + ),
  74.                     ans = max(ans, f[r][j+]);
  75.             }
  76.         }
  77.         printf("Case #%d: %d\n", kase, ans);
  78.     }
  79.     return ;
  80. }

Problem I - Inkopolis 04：59：59 （-3） Solved by Dancepted & xk

中午还在看的基环树下午就用到了。

先考虑树上的情况。会影响答案的部分只有修改颜色之前的颜色$color_{pre}$，和之后的颜色$color_{now}$两种。看是否会新增、删除color region，或者合并、分割原来的color region。

再考虑环上的情况。同上。特别地：如果整个环都是同一种颜色，修改了颜色不会分割原来的color region；如果整个环除了修改的边都是$color_{now}$，则修改颜色不会合并color region。

计算节日开始前各街道的颜色，可以模仿修改的操作，一条条边加入就行了。

实现的时候先把环扣出来，统计一下环各颜色的边数量。用个map保存每个点相邻的各颜色的数量。再用map保存每条边的颜色。

然后扫一遍所有的边先计算初始情况下的region的数量sum，然后一次次修改边的颜色，更新sum就好了。

代码：O(nlogn)

  1. #include <iostream>
  2. #include <cmath>
  3. #include <map>
  4. #include <algorithm>
  5. #include <cstdio>
  6. #include <cstring>
  7. #include <set>
  8. #include <vector>
  9. #include <string>
  10. #include <queue>
  11. #include <stack>
  12. #include <iomanip>
  13. #define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
  14. #define N 200005
  15. #define M 200005
  16. #define INF 0x3f3f3f3f
  17. #define mk(x) (1<<x) // be conscious if mask x exceeds int
  18. #define sz(x) ((int)x.size())
  19. #define upperdiv(a,b) (a/b + (a%b>0))
  20. #define mp(a,b) make_pair(a, b)
  21. #define endl '\n'
  22. #define lowbit(x) (x&-x)
  23.  
  24. using namespace std;
  25. typedef long long ll;
  26. typedef double db;
  27.  
  28. /** fast read **/
  29. template <typename T>
  30. inline void read(T &x) {
  31.     x = ; T fg = ; char ch = getchar();
  32.     while (!isdigit(ch)) {
  33.         if (ch == '-') fg = -;
  34.         ch = getchar();
  35.     }
  36.     while (isdigit(ch)) x = x*+ch-'', ch = getchar();
  37.     x = fg * x;
  38. }
  39. template <typename T, typename... Args>
  40. inline void read(T &x, Args &... args) { read(x), read(args...); }
  41. template <typename T>
  42. inline void write(T x) {
  43.     int len = ; char c[]; if (x < ) putchar('-'), x = -x;
  44.     do{++len; c[len] = x% + '';} while (x /= );
  45.     for (int i = len; i >= ; i--) putchar(c[i]);
  46. }
  47. template <typename T, typename... Args>
  48. inline void write(T x, Args ... args) { write(x), write(args...); }
  49.  
  50. int tot;
  51. int head[N], nxt[N<<], ver[N<<], col[N<<];
  52. void addEdge(int u, int v, int c) {
  53.     nxt[tot] = head[u], ver[tot] = v, col[tot] = c, head[u] = tot++;
  54. }
  55.  
  56. int iscir[N], cirlen;
  57. bool vis[N];
  58. int getCir(int u, int fa, int& rt) {
  59.     vis[u] = true;
  60.     for (int i = head[u]; i != -; i = nxt[i]) {
  61.         int v = ver[i];
  62.         if (v == fa) continue;
  63.         if (vis[v]) {
  64.             rt = v;
  65.             iscir[u] = ;
  66.             cirlen++;
  67.             return ;
  68.         }
  69.         else {
  70.             int tmpcir = getCir(v, u, rt);
  71.             if (tmpcir) {
  72.                 iscir[u] = tmpcir;
  73.                 cirlen++;
  74.                 return u != rt;
  75.             }
  76.         }
  77.     }
  78.     return ;
  79. }
  80.  
  81. int cnt[N];
  82. int sum;
  83. map<int, int> mp[N];
  84. map<pair<int,int>, int> uv_col;
  85.  
  86. void update(int u, int v, int ccur, bool ifprint) {
  87.     if (u > v) swap(u, v);
  88.     int cpre = uv_col[mp(u, v)];
  89.     if (cpre == ccur) {
  90.         if (ifprint)
  91.             printf("%d\n", sum);
  92.         return;
  93.     }
  94.     if (!iscir[u] || !iscir[v]) {
  95.         //on tree
  96.         if (mp[u][cpre] >=  && mp[v][cpre] >= ) {
  97.             sum++;
  98.         }
  99.         if (mp[u][cpre] ==  && mp[v][cpre] == ) {
  100.             sum--;
  101.         }
  102.         if (mp[u][ccur] && mp[v][ccur]) {
  103.             sum--;
  104.         }
  105.         if (!mp[u][ccur] && !mp[v][ccur]) {
  106.             sum++;
  107.         }
  108.     }
  109.     else {
  110.         //on circle
  111.         if (mp[u][cpre] >=  && mp[v][cpre] >= ) {
  112.             if (cnt[cpre] != cirlen)
  113.                 sum++;
  114.         }
  115.         if (mp[u][cpre] ==  && mp[v][cpre] == ) {
  116.             sum--;
  117.         }
  118.         if (mp[u][ccur] && mp[v][ccur]) {
  119.             if (cnt[ccur] != cirlen-)
  120.                 sum--;
  121.         }
  122.         if (!mp[u][ccur] && !mp[v][ccur]) {
  123.             sum++;
  124.         }
  125.  
  126.         if (cpre != -)
  127.             cnt[cpre]--;
  128.         cnt[ccur]++;
  129.     }
  130.     uv_col[mp(u, v)] = ccur;
  131.     if (cpre != -) {
  132.         mp[u][cpre]--;
  133.         mp[v][cpre]--;
  134.     }
  135.     mp[u][ccur]++;
  136.     mp[v][ccur]++;
  137.     if (ifprint)
  138.         printf("%d\n", sum);
  139. }
  140.  
  141. int main() {
  142.     int T;
  143.     cin >> T;
  144.     for (int kase = ; kase <= T; kase++) {
  145.         int n, m; read(n, m);
  146.         // init
  147.         uv_col.clear();
  148.         for (int i = ; i <= n; i++) {
  149.             mp[i].clear();
  150.         }
  151.         tot = ;
  152.         memset(head, -, (n+) * sizeof(int));
  153.  
  154.         // input and count colors on every edge
  155.         for (int i = ; i <= n; i++) {
  156.             int u, v, c; read(u, v, c);
  157.             addEdge(u, v, c);
  158.             addEdge(v, u, c);
  159.             if (u > v)
  160.                 swap(u, v);
  161.             uv_col[mp(u, v)] = -;
  162.             // mp[u][c]++; mp[v][c]++;
  163.         }
  164.         // get circle
  165.         int rt = -;
  166.         memset(vis, false, (n+) * sizeof(bool));
  167.         memset(iscir, , (n+) * sizeof(int));
  168.         cirlen = ;
  169.         getCir(, -, rt);
  170.         // count colors on circle
  171.         memset(cnt, , (n+) * sizeof(int));
  172.         sum = ;
  173.         for (int i = ; i < n; i++) {
  174.             int u = ver[i<<], v = ver[i<<|], ccur = col[i<<];
  175.             update(u, v, ccur, false);
  176.         }
  177.  
  178.         printf("Case #%d:\n", kase);
  179.         for (int i = ; i <= m; i++) {
  180.             int u, v, ccur; read(u, v, ccur);
  181.             update(u, v, ccur, true);
  182.         }
  183.     }
  184.     return ;
  185. }
  186. /*
  187. 11111
  188. 5 10
  189. 1 5 1
  190. 2 5 1
  191. 3 5 1
  192. 4 5 1
  193. 1 2 1
  194. 1 2 1
  195.  
  196. 2
  197. 4 4
  198. 1 2 1
  199. 2 3 1
  200. 3 4 1
  201. 4 1 1
  202. 1 2 2
  203. 3 4 2
  204. 2 3 2
  205. 4 1 4
  206. 4 3
  207. 4 2 4
  208. 2 3 3
  209. 3 4 2
  210. 1 4 1
  211. 3 4 2
  212. 2 3 4
  213. 3 4 3
  214. */

总结：

本场能绝杀还是有点运气成分在的，而且现场赛的话最后几分钟可能就交不上题了也可能。不过绝杀也不影响苟在银牌，问题不大。

然后最近的几场好像都是我在贡献罚时，感觉码力出现了一些小小的问题。这两天要小心一点。

然后还有两周可能就要退役了，模拟赛还没有进过金区感觉很难受啊，打成这个样子还怎么冲金啊qaq。

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