POJ2762Going from u to v or from v to u?

Going from u to v or from v to u?

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 13294 Accepted: 3460

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

The first line contains a single integer T, the number of test cases. And followed T cases. 
The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly. 

Output

The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.

Sample Input

13 31 22 33 1

Sample Output

Yes

题意：有N个山洞m条路，问任意两点x y 能否存在从x到y 或者从y到x。

思路：先对图求强连通分量，为每个连通块标号（如果有多个强连通分量），然后建缩点之后的图，图，即使代码中的G3，因为不同连通块间的标号不同，所以可以据此建图，G3就是一个DAG。

然后找到入度为零的点，如果个数大于等于2则这两个点无法连通（想想为什么），如果个数为1则进行

DAG上求最长路，如果最长路等于强连通的个数，则成立，否者条件不满足任意两点连通。

具体可以参考代码后的测试用数据。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <stack>#define DEBUGusing namespace std;const int maxn = 1000+10;vector<int> G[maxn], G2[maxn], G3[maxn];vector<int> S;bool flag = false;int vis[maxn];  // for dfs1();int sccno[maxn], scc_cnt;bool mark[maxn];// for Track_path.int d[maxn]; // record the length of every vertex  (DAG). for Track_path().int gn, gm;void dfs1(int u) {    if(vis[u]) return;    vis[u] = 1;    for(int i = 0; i < (int)G[u].size(); i++) dfs1(G[u][i]);    S.push_back(u);}void dfs2(int u) {    if(sccno[u]) return;    sccno[u] = scc_cnt;    for(int i = 0; i < (int)G2[u].size(); i++) dfs2(G2[u][i]);}void find_scc(int n) {    scc_cnt = 0;    flag = false;    S.clear();    memset(sccno, 0, sizeof(sccno));    memset(vis, 0, sizeof(vis));    for(int i = 1; i <= gn; i++) dfs1(i);    for(int i = n-1; i >= 0; i--) {        if(!sccno[S[i]]) {            scc_cnt++;            dfs2(S[i]);        }    }}void build_map() {    for(int i = 0; i < maxn; i++) {        G[i].clear(); G2[i].clear();    }    int u, v;    for(int i = 1; i <= gm; i++) {        scanf("%d%d", &u, &v); // directed graph.        G[u].push_back(v);        G2[v].push_back(u);    }}#ifndef DEBUGvoid mid_res() {    for(int i = 1; i <= gn; i++) {        printf("sccno[%d] = %d\n", i, sccno[i]);    }    cout << endl;}#endifvoid build_map2() {// build DAG。    for(int i = 0; i < maxn; i++) G3[i].clear();    for(int i = 1; i <= gn; i++) {        for(int j = 0; j < (int)G[i].size(); j++) {            if(sccno[i] != sccno[G[i][j]]) {                int u = sccno[i], v = sccno[G[i][j]];                G3[u].push_back(v);            }        }    }#ifndef   DEBUG    for(int i = 1; i <= scc_cnt; i++) {        for(int j = 0; j < (int)G3[i].size(); j++) {            printf("%d --> %d\n", i, G3[i][j]);        }    }#endif}void Track_path(int u) {    if(mark[u]) return;    mark[u] = true;    for(int i = 0; i < (int)G3[u].size(); i++) {        int v = G3[u][i];        if(!mark[v]) {            d[v] = d[u] + 1;            Track_path(v);        }    }}void work() {    bool tag[maxn];    memset(tag, false, sizeof(tag));    for(int i = 1; i <= gn; i++) {        if(tag[sccno[i]]) continue; // outdegree != 0.        for(int j = 0; j < (int)G2[i].size(); j++) {            if(sccno[i] != sccno[G2[i][j]]) {                tag[sccno[i]] = true; //  reverse graph outdegree > 0.                break;            }        }    }    vector<int> ss;    ss.clear();    for(int i = 1; i <= scc_cnt; i++) {        if(tag[i] == false) {            ss.push_back(i);          //  printf("scc_cnt = %d\n", i);        }    }    if(ss.size() > 1) {        flag = false;        return;    }    build_map2();    memset(mark, false, sizeof(mark));    memset(d, 0, sizeof(d));    d[ss[0]] = 1;    Track_path(ss[0]);    int max_length = -1;    for(int i = 1; i <= scc_cnt; i++) {        max_length = max(max_length, d[i]);    }    if(max_length == scc_cnt) flag = true;    return;}int main(){    int T;    scanf("%d", &T);    while(T--) {        scanf("%d%d", &gn, &gm);        build_map();        find_scc(gn);       // mid_res();        work();        if(flag) printf("Yes\n");        else printf("No\n");    }    return 0;}/**49 101 22 33 12 98 99 44 55 66 77 44 41 21 32 43 43 31 22 33 1**/