A:签到题

C:模拟搜索题

#include <bits/stdc++.h>

#include <cstring>

#include <iostream>

#include <algorithm>

#include<queue>

#define EPS 1.0e-9

#define PI acos(-1.0)

#define INF 30000000

#define MOD 1000000007

#define mem(a,b) memset((a),b,sizeof(a))

#define TS printf("!!!\n")

#define pb push_back

#define pai pair<int,int>

//using ll = long long;

//using ull= unsigned long long;

//std::ios::sync_with_stdio(false);

using namespace std;

//priority_queue<int,vector<int>,greater<int>> que;

typedef long long ll;

typedef unsigned long long ull;

const int maxn=;

int year[];

int where[];

int visit[];

vector<int> manager[];

int anser=;

void dfs(int x)

{

visit[x]=;

int len=manager[x].size();

for(int i=;i<len;i++)

{

int to=manager[x][i];

if(!visit[to])

{

anser=min(anser,year[to]);

dfs(to);

}

}

}

int main()

{

        for(int i=;i<=;i++)

        where[i]=i;

        int n,m,I;

        int from,to;

        cin >> n >>m >>I;

        for(int i=;i<=n;i++)

        cin >>year[i];

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

        {

        scanf("%d %d",&from,&to);

        manager[to].pb(from);

        }

        string a;

        int cnt;

        for(int i=;i<=I;i++)

        {

        cin >> a;

        if(a[]=='T')

        {

        cin >> from >>to;

        swap(where[from],where[to]);

        swap(year[where[from]],year[where[to]]);

        }

        else if(a[]=='P')

        {

        mem(visit,);

        cin >> from;

        anser=;

        dfs(where[from]);

        if(anser==)

        cout<<"*"<<endl;

        else

        cout<<anser<<endl;

        }

        }

}

D:队友

#include<stdio.h>

#include<math.h>

#include<string.h>

#include<iostream>

#include<algorithm>

#include<map>

#define INF 2139062143

#define inf -2139062144

#define ll long long

#include<vector>

using namespace std;

map<vector<int> , int> ma;

vector<int> rr;

bool jj = false;

vector<int> fold(int p,vector<int> qq) {

    int l = qq.size();

    vector<int> re;

    for(int i=; i<=l; i++) {

        int tt = ;

        bool flag = false;

        if(p - i >= ) {

            tt+=qq[p-i];

            flag = true;

        }

        if(p + i -  < l) {

            tt+=qq[p + i - ];

            flag = true;

        }

        if(flag)    re.insert(re.begin(),tt);

    }

    return re;

}

void dfs(vector<int> qq) {

    int llen = qq.size();

    int i,j;

    for(i=; i<=llen && jj == false; i++) {

        vector<int> tem = fold(i,qq);

        if(tem == rr) {

            jj = true;

            break;

        }

        if(ma[tem] == ) {

            ma[tem] = ;

            dfs(tem);

        }

    }

}

int main() {

    vector<int> q;

    int n,i,j;

    scanf("%d",&n);

    for(i=; i<n; i++) {

        int a;

        scanf("%d",&a);

        q.push_back(a);

    }

    int m;

    scanf("%d",&m);

    for(i=; i<m; i++) {

        int a;

        scanf("%d",&a);

        rr.push_back(a);

    }

    dfs(q);

//    for(i=0; i<=n; i++) {

//        vector<int> tem = fold(i);

//        for(j=0; j<tem.size(); j++) {

//            printf("%d ",tem[j]);

//        }

//        printf("\n");

//        if(tem == rr) {

//            jj = true;

//            break;

//        }

//    }

    if(jj)    printf("S\n");

    else    printf("N\n");

    return ;

}

E:水题

#include <bits/stdc++.h>

#include <cstring>

#include <iostream>

#include <algorithm>

#include<queue>

#define EPS 1.0e-9

#define PI acos(-1.0)

#define INF 30000000

#define MOD 1000000007

#define mem(a,b) memset((a),b,sizeof(a))

#define TS printf("!!!\n")

#define pb push_back

#define pai pair<int,int>

//using ll = long long;

//using ull= unsigned long long;

//std::ios::sync_with_stdio(false);

using namespace std;

//priority_queue<int,vector<int>,greater<int>> que;

typedef long long ll;

typedef unsigned long long ull;

const int maxn=;

int a[];

int visit[];

queue<int> que;

int main()

{

      mem(visit,);

      int r,n;

      cin >>r >> n;

      int now;

      for(int i=;i<=n;i++)

      {

        scanf("%d",&now);

        visit[now]=;

      }

      for(int i=;i<=r;i++)

        if(!visit[i])

        que.push(i);

      int len=que.size();

      if(len==)

      {

        cout<<"*"<<endl;

        return ;

      }

      while(!que.empty())

      {

        cout<<que.front()<<" ";

        que.pop();

      }

      cout<<endl;

}

F:队友

#include<stdio.h>

#include<math.h>

#include<string.h>

#include<iostream>

#include<algorithm>

#include<map>

#define INF 2139062143

#define inf -2139062144

#define ll long long

using namespace std;

int a[];

int p[];

map<int,int> q;

int main(){

    int n,i,j;

    scanf("%d",&n);

    int sum = ;

    for(i=;i<=n;i++){

        scanf("%d",&a[i]);

        sum += a[i];

        p[i] = sum;

        q[sum] = ;

//        printf("%d ",sum);

    }

//    printf("\n");

    a[] = ;

    q[] = ;

    p[] = ;

    if(sum %  != ){

        printf("0\n");

        return ;

    }

    int len = sum / ;

    int ans = ;

    for(i = ;i < n;i++){

//        printf("%d %d %d\n",p[i],(p[i] + len) % sum,(p[i] + len * 2) % sum);

        if(q[(p[i] + len) % sum] ==  && q[(p[i] + len * ) % sum] == ){

            ans++;

        }

    }

    printf("%d\n",ans/);

    return ;

}

G:队友

#include <bits/stdc++.h>

using namespace std;

struct Node {

    int val, id;

    bool operator < (const Node & rhs) const {

        return val < rhs.val;

    }

}a[];

int L, C, X[][], cnt_l[], cnt_c[], l[], c[], b[];

int Solve(int * array, int n) {

    for(int i = ; i <= n; i++) a[i].val = array[i], a[i].id = i;

    sort(a + , a +  + n);

    for(int i = ; i <= n; i++) b[i] = a[i].id;

    int ans = ;

    for(int i = ; i <= n ;i++) {

        if(b[i] != i) {

            for(int j = i; j <= n; j++) {

                if(b[j] == i) {

                    swap(b[j], b[i]);

                    ans++;

                    break;

                }

            }

        }

    }

    return ans;

}

int main() {

    scanf("%d%d", &L, &C);

    for(int i = ; i <= L; i++) {

        for(int j = ; j <= C; j++) {

            scanf("%d", &X[i][j]);

        }

    }

    int flag1 = , flag2 = , cnt, sum;

    for(int i = ; i <= L; i++) {

        cnt_l[(i - ) * C * C + ( + C) * C / ] = ;

        //cout << (i - 1) * C * C + (1 + C) * C / 2 << endl;

    }

    for(int i = ; i <= C; i++) {

        cnt_c[i * L + C * (L - ) * L / ] = ;

        //cout << i * L + C * (L - 1) * L / 2 << endl;

    }

    cnt = ;

    for(int i = ; i <= L; i++) {

        sum = ;

        for(int j = ; j <= C; j++) {

            sum += X[i][j];

        }

        //cout << sum << endl;

        if(cnt_l[sum]) cnt++;

    }

    //cout << cnt << endl;

    if(cnt == L) flag1 = ;

    cnt = ;

    for(int j = ; j <= C; j++) {

        sum = ;

        for(int i = ; i <= L; i++) {

            sum += X[i][j];

        }

        if(cnt_c[sum]) cnt++;

    }

    if(cnt == C) flag2 = ;

    //cout << flag1 << " " << flag2 << endl;

    if(flag1 && flag2) {

        for(int i = ; i <= L; i++) {

            l[i] = X[i][];

        }

        for(int i = ; i <= C; i++) {

            c[i] = X[][i];

        }

        int ans = ;

        ans += Solve(l, L);

        ans += Solve(c, C);

        printf("%d\n", ans);

    }else {

        printf("*\n");

    }

    return ;

}

H:数论题 矩阵快速幂

#include<stdio.h>

#include<math.h>

#include<string.h>

#include<iostream>

#include<algorithm>

#define INF 2139062143

#define inf -2139062144

#define ll long long

using namespace std;

ll mod = ;

struct matrix {

    ll sizex,sizey;

    ll num[][];

};

matrix mul(matrix a,matrix b) {

    matrix result;

    result.sizex = a.sizex;

    result.sizey = b.sizey;

    ll i,j,k;

    for(i=; i<a.sizex; i++) {

        for(j=; j<b.sizey; j++) {

            ll sum = ;

            for(k=; k<a.sizey; k++)    sum+=a.num[i][k]*b.num[k][j] % mod;

            result.num[i][j] = sum % mod;

        }

    }

    return result;

}

matrix zero(ll size) {

    ll i,j;

    matrix result;

    result.sizex = size;

    result.sizey = size;

    for(i=; i<size; i++)

        for(j=; j<size; j++)

            result.num[i][j] = ;

    return result;

}

matrix eye(ll size) {

    matrix result;

    result.sizex = size;

    result.sizey = size;

    result = zero(size);

    for(ll i=; i<size; i++) {

        result.num[i][i] = ;

    }

    return result;

}

matrix quickmat(matrix a,ll k) {

    matrix origin = a;

    matrix res;

    res = eye(a.sizex);

    while(k) {

        if (k & ) {

            res=mul(res,origin);

        }

        k>>=;

        origin=mul(origin,origin);

    }

    return res;

}

matrix initial(ll *a,ll sizex,ll sizey) {

    ll i,j;

    matrix result;

    result.sizex = sizex;

    result.sizey = sizey;

    ll count = ;

    for(i=; i<sizex; i++)

        for(j=; j<sizey; j++)

            result.num[i][j] = a[count++];

    return result;

}

ll all_ele_sum(matrix a) {

    ll re = ;

    for(ll i = ; i<a.sizex; i++) {

        for(ll j = ; j<a.sizey; j++) {

            re+=a.num[i][j];

        }

    }

    return re;

}

void printmatrix(matrix a) {

    for(ll i = ; i<a.sizex; i++) {

        for(ll j = ; j<a.sizey; j++) {

            printf("%lld    ",a.num[i][j]);

        }

        printf("\n");

    }

    printf("\n");

}

int main() {

    ll n,k,l;

    scanf("%lld%lld%lld",&n,&k,&l);

    n /= ;

    k %= mod;

    l %= mod;

    ll a[] = {k,l,,};

    if(n == ){

        printf("%06lld\n",a[]);

        return ;

    }

    ll x2 = (k * k % mod + l) % mod;

    if(n == ){

        printf("%06lld\n",x2);

        return ;

    }

    matrix temp = initial(a,,);

    matrix re = quickmat(temp,n-);

    ll ans = (re.num[][] * x2 % mod + re.num[][] * k % mod) % mod;

    printf("%06lld\n",ans);

    return ;

}

I:类搜 DP

#include <bits/stdc++.h>

#include <cstring>

#include <iostream>

#include <algorithm>

#include<queue>

#define EPS 1.0e-9

#define PI acos(-1.0)

#define INF 30000000

#define MOD 1000000007

#define mem(a,b) memset((a),b,sizeof(a))

#define TS printf("!!!\n")

#define pb push_back

#define pai pair<int,int>

//using ll = long long;

//using ull= unsigned long long;

//std::ios::sync_with_stdio(false);

using namespace std;

//priority_queue<int,vector<int>,greater<int>> que;

typedef long long ll;

typedef unsigned long long ull;

const int maxn=;

int f[maxn];

int n;

int c;

int t1,t2;

int tmax,tmin;

ll ans[maxn];

ll anser;

ll get(int x)

{

 if(x>n)

 return ;

 if(ans[x]!=-)

 return ans[x];

 ans[x]=<<;

 for(int i=x;i<=n;i++)

 {

        ll cha=f[i]-f[x];

        if(cha>tmax)

        break;

        if(cha<=t1)

        ans[x]=min(ans[x],get(i+)+t1);

        if(cha<=t2)

        ans[x]=min(ans[x],get(i+)+t2);

 }

 return ans[x];

}

int main()

{

        mem(ans,-);

        cin >> n >> c >> t1 >> t2;

        tmax=max(t1,t2);

        tmin=min(t1,t2);

        for(int i=;i<=n;i++)

        cin >> f[i];

        sort(f+,f++n);

        ll anser=get();

        cout<<anser<<endl;

}

J:LCA/LCT

#include <bits/stdc++.h>

#define MAXN 20005

#define MAXM 100005

#define inf 0x3f3f3f3f

using namespace std;

int n, m, s;

int head[MAXN], cnt;

int fa[MAXN];

int ancient[MAXN][], minw[MAXN][], dep[MAXN];

struct Node {

    int u, v, w;

    bool operator<(const Node &rhs) const {

        return w > rhs.w;

    }

} ori_edge[MAXM];

struct Edge {

    int to, nxt, w;

} edge[MAXM * ];

int Find(int x) {

    return fa[x] == x ? x : fa[x] = Find(fa[x]);

}

void AddEdge(int u, int v, int w) {

    edge[cnt].to = v;

    edge[cnt].w = w;

    edge[cnt].nxt = head[u];

    head[u] = cnt++;

}

void Add(int u, int v, int w) {

    AddEdge(u, v, w);

    AddEdge(v, u, w);

}

void Init() {

    memset(head, -, sizeof(head));

    cnt = ;

    for (int i = ; i <= n; i++) fa[i] = i;

}

void Kruskal() {

    sort(ori_edge + , ori_edge + m + );

    int cnt = ;

    for (int i = ; i <= m; i++) {

        int f1 = Find(ori_edge[i].u);

        int f2 = Find(ori_edge[i].v);

        if (f1 != f2) {

            fa[f1] = f2;

            Add(ori_edge[i].u, ori_edge[i].v, ori_edge[i].w);

            cnt++;

            if (cnt == n - ) break;

        }

    }

}

void Dfs(int u, int f, int d) {

    fa[u] = f;

    dep[u] = d;

    for (int i = head[u]; ~i; i = edge[i].nxt) {

        int v = edge[i].to;

        if (v == f) continue;

        minw[v][] = edge[i].w;

        Dfs(v, u, d + );

    }

}

void Pre() {

    memset(ancient, -, sizeof(ancient));

    for (int i = ; i <= n; i++) {

        ancient[i][] = fa[i];

    }

    minw[][] = inf;

    for (int j = ; ( << j) <= n; j++) {

        for (int i = ; i <= n; i++) {

            if (~ancient[i][j - ]) {

                int f = ancient[i][j - ];

                ancient[i][j] = ancient[f][j - ];

                minw[i][j] = min(minw[i][j - ], minw[f][j - ]);

            }

        }

    }

}

int Query(int p, int q) {

    if (dep[p] < dep[q]) swap(p, q);

    int k = ;

    while (( << k) < dep[p]) k++;

    int ans = inf;

    for (int i = k; i >= ; i--) {

        if (~ancient[p][i] && dep[ancient[p][i]] >= dep[q]) {

            ans = min(ans, minw[p][i]), p = ancient[p][i];

        }

    }

    if (p == q) return ans;

    for (int i = k; i >= ; i--) {

        if (~ancient[p][i] && ancient[p][i] != ancient[q][i]) {

            ans = min(ans, minw[p][i]);

            p = ancient[p][i];

            ans = min(ans, minw[q][i]);

            q = ancient[q][i];

        }

    }

    if (p != q) {

        ans = min(ans, minw[p][]);

        ans = min(ans, minw[q][]);

    }

    return ans;

}

int main() {

    scanf("%d%d%d", &n, &m, &s);

    Init();

    for (int i = ; i <= m; i++) {

        scanf("%d%d%d", &ori_edge[i].u, &ori_edge[i].v, &ori_edge[i].w);

    }

    Kruskal();

    Dfs(, -, );

    Pre();

    for (int i = ; i <= s; i++) {

        int l, h;

        scanf("%d%d", &l, &h);

        printf("%d\n", Query(l, h));

    }

    return ;

}

Summer training round2 #9(Training28)的更多相关文章

  1. Summer training round2 #3

    A!:                    GTY系列题 B!:莫队加分块  GTY系列题 C!:线段树模拟拓扑排序(把普通的拓扑排序的栈操作改成线段树区间减一,查询区间最右侧的0的位置即可.注意一 ...

  2. Summer training round2 #1

    A:水 B:求两个三角形之间的位置关系:相交 相离 内含 ①用三个点是否在三角形内外判断    计算MA*MB.MB*MC.MC*MA的大小 若这三个值同号,那么在三角形的内部,异号在外部 #incl ...

  3. Summer training round2 #8(Training26)

    A:贪心DFS 先从最远的搜起 如果一个点的value>=2 就ans++ D:并查集 E:大模拟 F:快速幂 #include <bits/stdc++.h> using name ...

  4. Summer training round2 #10(Training 30)

    A:签到题 B!:搜索+DP #include<bits/stdc++.h> #define mp make_pair #define pi pair<int,int> usi ...

  5. Summer training round2 #7 (Training #23)

    A:约瑟夫环 套公式 B:线性筛素数 C:投骰子 概率DP F:有权无向图的生成树(边最大值和最小值只差最小) 直接kruskal G:状压BFS或者双向BFS H:模拟题 I:几何题 J:高斯消元

  6. Summer training round2 #6 (Training #22)

    A:二分答案 如果中位数比目前的大就right=mid-1 else left=mid+1 C!:几何 G:优先队列贪心 #include <bits/stdc++.h> using na ...

  7. Summer training round2 #5 (Training #21)

    A:正着DFS一次处理出每个节点有多少个优先级比他低的(包括自己)作为值v[i] 求A B 再反着DFS求优先级比自己高的求C #include <bits/stdc++.h> #incl ...

  8. Summer training round2 #4 (Training #20)

    A!:UESTC1752 B!:找区间内L到R之间内的数的个数  权值分块加莫队 C!:给你一个哈斯图 去掉其中的几条边 要求输出字典序最大的拓扑排序:线段树模拟拓扑排序 D!:要求你找到最短路树并输 ...

  9. hdu 4946 2014 Multi-University Training Contest 8

    Area of Mushroom Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) ...

随机推荐

  1. Client Dimensions , offsetHeight , scrollTop 属性详解

    http://stackoverflow.com/questions/22675126/what-is-offsetheight-clientheight-scrollheight http://ww ...

  2. PHP xdebug 断点调试

    转载自: https://blog.csdn.net/qq_32631847/article/details/82054011

  3. 【0.1】mysql版本升级（5.6升级到5.7）

    目录 [1].升级操作 [2].mysql 5.6安装(二进制) [3].mysql 5.7安装(二进制) [1].升级操作 核心步骤 [1.1]停止mysql 5.6 [1.2]把环境变量引用到My ...

  4. GrapeCity Documents for Excel 与 Apache POI 功能对比

    GrapeCity Documents for Excel 与 Apache POI 功能对比 GrapeCity Documents for Excel 是什么? GrapeCity Documen ...

  5. Shell脚本编程（一）

    shell 脚本编程(一) 1 . shell 的作用 Shell的作用是解释执行用户的命令,用户输入一条命令,Shell就解释执行一条,这种方式称为交互式(Interactive),Shell还有一 ...

  6. 注入(Injection)

    注入(Injection)是: Java EE提供了注入机制,使您的对象能够获取对资源和其他依赖项的引用,而无需直接实例化它们.通过使用将字段标记为注入点的注释之一来装饰字段或方法,可以在类中声明所需 ...

  7. Hive 教程(三)-DDL基础

    DDL,Hive Data Definition Language,数据定义语言: 通俗理解就是数据库与库表相关的操作,本文总结一下基本方法 hive 数据仓库配置 hive 数据仓库默认位置在 hd ...

  8. L2-013. 红色警报(并查集+无向图联通分量)

    战争中保持各个城市间的连通性非常重要.本题要求你编写一个报警程序,当失去一个城市导致国家被分裂为多个无法连通的区域时,就发出红色警报.注意:若该国本来就不完全连通,是分裂的k个区域,而失去一个城市并不 ...

  9. CentOS 7系统安装Jenkins

    一.jenkins 介绍 Jenkins是一个开源的.可扩展的持续集成.交付.部署(软件/代码> 的编译.打包.部署)基于web界面的平台. 简单说,就是各种项目的"自动化" ...

  10. leecode刷题（28）-- 二叉树的前序遍历

    leecode刷题(28)-- 二叉树的前序遍历 二叉树的前序遍历 给定一个二叉树,返回它的 前序 遍历. 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,2,3] 思路 ...