NOIP2010 题解

机器翻译

　　题解：模拟

 #include <cstdio>
 #include <cstring>

 const int MAXN = ;

 int n, m, cj, now, MinV, MinId, Pri, a[MAXN+], in[MAXN+];

 int main(){
     memset(in, -, sizeof(in));
     scanf("%d %d", &n, &m), now = Pri = ;
     for (int i=; i<m; i++){
         scanf("%d", &cj);
         if (~in[cj]) continue;
         Pri ++;
         if (now+<=n) now ++, in[cj] = i;
         else {
             MinV = 0x3f3f3f3f;
             for (int j=; j<=MAXN; j++)
                 if (in[j]<MinV && in[j]!=-) MinV = in[j], MinId = j;
             in[MinId] = -, in[cj] = i;
         }
     }
     printf("%d\n", Pri);
 }

translate.cpp

乌龟棋

　　题解：dp

 #include <cstdio>
 #include <cstring>

 const int MAXN = +;
 const int MAXE = +;

 int n, m, cj, a[MAXN], b[], dp[MAXE][MAXE][MAXE][MAXE];

 int main(){
     memset(dp, , sizeof(dp));

     scanf("%d %d", &n, &m);
     for (register int i=; i<n; i++)
         scanf("%d", &a[i]);
     for (register int i=; i<m; i++)
         scanf("%d", &cj), b[cj] ++;

     dp[][][][] = a[];
     for (register int A=; A<=b[]; A++)
         for (register int B=; B<=b[]; B++)
             for (register int C=; C<=b[]; C++)
                 for (register int D=; D<=b[]; D++){
                     cj = A+B+B+C+C+C+D+D+D+D;
                     if (A && dp[A-][B][C][D]+a[cj]>dp[A][B][C][D])    dp[A][B][C][D] = dp[A-][B][C][D]+a[cj];
                     if (B && dp[A][B-][C][D]+a[cj]>dp[A][B][C][D]) dp[A][B][C][D] = dp[A][B-][C][D]+a[cj];
                     if (C && dp[A][B][C-][D]+a[cj]>dp[A][B][C][D]) dp[A][B][C][D] = dp[A][B][C-][D]+a[cj];
                     if (D && dp[A][B][C][D-]+a[cj]>dp[A][B][C][D]) dp[A][B][C][D] = dp[A][B][C][D-]+a[cj];
                 }
     printf("%d\n", dp[b[]][b[]][b[]][b[]]);
 }

tortoise.cpp

关押罪犯

　　题解：并查集。既然只有两个监狱，那么如果a和b没有在一个监狱，b和c没有在一个监狱，那么a和c一定在一个监狱

　　把怒气值从大到小拍个序，依次处理

　　并查集[1, n]表示和自己一起的，[n+1, 2n]表示不和自己一起的

 #include <cstdio>
 #include <algorithm>
 using std::sort;

 const int MAXN = +;
 const int MAXM = +;

 int n, m, f[MAXN*];

 struct Relation{
     int a, b, c;

     friend bool operator < (const Relation& A, const Relation& B){
         return A.c>B.c;
     }
 }r[MAXM];

 inline int Find(int x){
     return f[x]==x ? x : f[x] = Find(f[x]);
 }

 inline int Solve(){
     sort(r, r+m);
     int a, b;
     for (int i=; i<m; i++){
         a = Find(r[i].a), b = Find(r[i].b);
         if (a==b) return r[i].c;
         else f[a] = Find(r[i].b+n), f[b] = Find(r[i].a+n);
     }
     return ;
 }

 int main(){
     scanf("%d %d", &n, &m);
     for (int i=; i<=*n; i++)
         f[i] = i;
     for (int i=; i<m; i++)
         scanf("%d %d %d", &r[i].a, &r[i].b, &r[i].c);

     printf("%d\n", Solve());
 }

prison.cpp

引水入城

　　题解：看懂题是关键...

　　　　　多么痛的领悟：windows下只能bfs，linux下就怎么搞都可以了TAT

 #include <cstdio>
 #include <cstring>

 const int MAXN = +;
 const int addX[] = {-, , , };
 const int addY[] = {, , -, };

 int n, m, X, Y, Need, Cant, anna[MAXN][MAXN];
 bool vis[MAXN][MAXN];

 struct Region{
     int l, r;
 }rg[MAXN];

 inline int findLeft(){
     for (int i=; i<=m; i++)
         if (vis[n][i]) return i;
     return ;
 }

 inline int findRight(){
     for (int i=m; i>=; i--)
         if (vis[n][i]) return i;
     return ;
 }

 inline void DFS(int x, int y){
     if (x< || x>n || y< || y>m) return;
     vis[x][y] = true;

     for (int i=; i<; i++){
         X = x+addX[i], Y = y+addY[i];
         if (<=X && X<=n && <=Y && Y<=m && anna[X][Y]<anna[x][y] && !vis[X][Y])
             DFS(X, Y);
     }
 }

 inline bool legal(){
     memset(vis, false, sizeof(vis));

     for (int i=; i<=m; i++)
         DFS(, i);

     for (int i=; i<=m; i++)
             if (not vis[n][i]) ++ Cant;

     return !Cant;
 }

 inline int segmentCover(){
     for (int i=; i<=m; i++){
         memset(vis, false, sizeof(vis));
         DFS(, i);
         rg[i].l = findLeft(), rg[i].r = findRight();
     }

     int i = , j, k = ;
     while (i<=m){
         for (j=; j<=m; j++)
             if (rg[j].l<=i && rg[j].r>k) k = rg[j].r;
         i = k+, ++ Need;
     }

     return Need;
 }

 int main(){
     memset(anna, 0x3f, sizeof(anna));

     scanf("%d %d", &n, &m), Cant = Need = ;
     for (int i=; i<=n; i++)
         for (int j=; j<=m; j++)
             scanf("%d", &anna[i][j]);

     if (legal()) printf("1\n%d", segmentCover());
     else printf("0\n%d", Cant);
 }

flow.cpp