poj2762 判断一个图中任意两点是否存在可达路径 也可看成DAG的最小覆盖点是否为1

 

Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17993		Accepted: 4816

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

1
3 3
1 2
2 3
3 1

Sample Output

Yes


题目大意就是给你张图，问你是否任意两点u,v 可以从u到v或者从v到u
两种解法 先贴拓扑的
另外有些数据
1 
3 2
1 2
3 2

1
3 3
1 2
1 3
2 3

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=1008;
const int M=6008;
int head[N],dfn[N],in[N],low[N],bl[N],q[N];
int tot,cnt,scnt,n,m,l;
bool instack[N];
bool adj[N][N];
struct node{
    int u,to,next;
}e[M<<1];
void init(){
    memset(dfn,0,sizeof(dfn));
    memset(instack,0,sizeof(instack));
    memset(in,0,sizeof(in));
    memset(head,-1,sizeof(head));
    memset(adj,0,sizeof(adj));
    l=tot=cnt=scnt=0;
}
void add(int u,int v){
    e[tot].u=u;e[tot].to=v;e[tot].next=head[u];head[u]=tot++;
}
void Tajan(int u){
    instack[u]=1;
    q[l++]=u;
    low[u]=dfn[u]=++cnt;
    for(int i=head[u];i+1;i=e[i].next){
        int v=e[i].to;
        if(!dfn[v]) {
            Tajan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(instack[v]&&low[u]>dfn[v]) low[u]=dfn[v];
    }
    if(low[u]==dfn[u]){
        int t;
        ++scnt;
        do{
            t=q[--l];
            instack[t]=0;
            bl[t]=scnt;
        }while(t!=u);
    }
}
bool Ju(int u){
    while(scnt--){
        int cont=0;
        for(int i=head[u];i+1;i=e[i].next){
            int v=e[i].to;
            --in[v];
            if(in[v]==0) {
                cont++;u=v;
            }
        }
        if(cont>1) return 0;
    }
    return 1;
}
int main(){
    int T,u,v;
    for(scanf("%d",&T);T--;){
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=m;++i){
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        for(int i=1;i<=n;++i)
            if(!dfn[i]) Tajan(i);
        memset(head,-1,sizeof(head));
        for(int i=0;i<tot;++i){
            u=e[i].u,v=e[i].to;
            if(bl[u]==bl[v]||adj[u][v]) continue;
            else {
                adj[bl[u]][bl[v]]=1; // 注意这里
                add(bl[u],bl[v]);
                ++in[bl[v]];
            }
        }
        int cont=0,k;
        for(int i=1;i<=scnt;++i)
            if(!in[i]) {++cont;k=i;}
        if(cont>1) {puts("No");continue;}
        else {
            if(Ju(k)) puts("Yes");
            else puts("No");
        }
    }
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int MAXN = 20010;
const int MAXM = 100010;

struct Edge{
    int to, next;
}edge[MAXM];

int head[MAXN], tot;
int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN];
int Index, top;
int scc;
bool Instack[MAXN];
int num[MAXN];
int n, m;

void init() {
    tot = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v) {
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

void Tarjan(int u) {
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        v = edge[i].to;
        if (!DFN[v]) {
            Tarjan(v);
            if (Low[u] > Low[v]) Low[u] = Low[v];
        }
        else if (Instack[v] && Low[u] > DFN[v])
            Low[u] = DFN[v];
    }
    if (Low[u] == DFN[u]) {
        scc++;
        do {
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc;
            num[scc]++;
        } while (v != u);
    }
}

void solve() {
    memset(Low, 0, sizeof(Low));
    memset(DFN, 0, sizeof(DFN));
    memset(num, 0, sizeof(num));
    memset(Stack, 0, sizeof(Stack));
    memset(Instack, false, sizeof(Instack));
    Index = scc = top = 0;
    for (int i = 1; i <= n; i++)
        if (!DFN[i])
            Tarjan(i);
}

vector<int> g[MAXN];
int linker[MAXN], used[MAXN];

bool dfs(int u) {
    for (int i = 0; i < g[u].size(); i++) {
        int v = g[u][i];
        if (!used[v]) {
            used[v] = 1;
            if (linker[v] == -1 || dfs(linker[v])) {
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}

int hungary() {
    int res = 0;
    memset(linker, -1, sizeof(linker));
    for (int i = 1; i <= scc; i++) {
        memset(used, 0, sizeof(used));
        if (dfs(i)) res++;
    }
    return (scc - res)==1;
}

int main() {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d%d", &n, &m);
        init();
        int u, v;
        for (int i = 0; i < m; i++) {
            scanf("%d%d", &u, &v);
            addedge(u, v);
        }
        solve();

        for (int i = 0; i <= scc; i++) g[i].clear();
        for (int u = 1; u <= n; u++) {
            for (int i = head[u]; i != -1; i = edge[i].next) {
                int v = edge[i].to;
                if (Belong[u] != Belong[v])
                    g[Belong[u]].push_back(Belong[v]);
            }
        }
        if(hungary()) puts("Yes");
        else puts("No");
    }
    return 0;
}