POJ 2762 Going from u to v or from v to u?

Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12260		Accepted: 3139

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

Source

POJ Monthly--2006.02.26,zgl & twb

 

 

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：注意是或 而不是和。所以缩点后，不是判断强连通为1。
第一种方法：从入度为0的点 DFS 搜索 最长能到达的路径 如果路径长度等于点数，说明可以从源点访问完所有点。   

 

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int VM=1010;
const int EM=6010;

struct Edge{
    int to,nxt;
}edge1[EM],edge2[EM];

int n,m,cnt,dep,top,atype,head1[VM],head2[VM];
int dfn[VM],low[VM],vis[VM],indeg[VM],outdeg[VM],belong[VM];
int stack[VM];

void Init(){
    cnt=0, dep=0, top=0, atype=0;
    memset(head1,-1,sizeof(head1));
    memset(head2,-1,sizeof(head2));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(vis,0,sizeof(vis));
    memset(indeg,0,sizeof(indeg));
    memset(outdeg,0,sizeof(outdeg));
    memset(belong,0,sizeof(belong));
}

void addedge(int cu,int cv,Edge edge[],int head[]){
    edge[cnt].to=cv;
    edge[cnt].nxt=head[cu];
    head[cu]=cnt++;
}

void Tarjan(int u){
    dfn[u]=low[u]=++dep;
    stack[top++]=u;
    vis[u]=1;
    for(int i=head1[u];i!=-1;i=edge1[i].nxt){
        int v=edge1[i].to;
        if(!dfn[v]){
            Tarjan(v);
            low[u]=min(low[u],low[v]);
        }else if(vis[v])
            low[u]=min(low[u],dfn[v]);
    }
    int j;
    if(dfn[u]==low[u]){
        atype++;
        do{
            j=stack[--top];
            belong[j]=atype;
            vis[j]=0;
        }while(j!=u);
    }
}

int path[VM],maxdep;

void DFS(int u,int d){
    path[u]=d;
    maxdep=max(maxdep,d);
    for(int i=head2[u];i!=-1;i=edge2[i].nxt){
        int v=edge2[i].to;
        if(path[u]>path[v])
            DFS(v,d+1);
    }
}

int Solve(){
    int u,v;
    for(u=1;u<=n;u++)
        for(int i=head1[u];i!=-1;i=edge1[i].nxt){
            v=edge1[i].to;
            if(belong[u]!=belong[v]){
                indeg[belong[v]]++;
                addedge(belong[u],belong[v],edge2,head2);
            }
        }
     maxdep=0;
     for(u=1;u<=n;u++)
        if(!indeg[u]){
            DFS(u,1);
            break;
        }
    if(maxdep==atype)
        return 1;
    return 0;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        Init();
        scanf("%d%d",&n,&m);
        int u,v;
        while(m--){
            scanf("%d%d",&u,&v);
            addedge(u,v,edge1,head1);
        }
        for(int i=1;i<=n;i++)
            if(!dfn[i])
                Tarjan(i);
        if(Solve())
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}

 

 

 

 

第二种方法： 只要用拓扑判断入度为0的点是否为1个，如果有多个，就出现在分叉，不可能从x到y。

 

 

#include <stdio.h>
#include <string.h>
#define VM 1005
#define EM 6005

struct E
{
    int to,next;
}edge[EM];
struct E1
{
    int frm,to,next;
}edge1[EM];
int head[VM],head1[VM],indeg[VM],map[VM][VM];
int dfn[VM],low[VM],vis[VM],belong[VM],stack[VM];
int scc,cnt,top,p,n;

void addedge (int cu,int cv)
{
    edge[p].to = cv;
    edge[p].next = head[cu];
    head[cu] = p++;
}

void addedge1(int cu,int cv)
{
    edge1[p].frm = cu;
    edge1[p].to = cv;
    edge1[p].next = head1[cu];
    head1[cu] = p ++;
}
void tarjan (int u)
{
    int v;
    dfn[u] = low[u] = ++cnt;
    stack[top++] = u;
    vis[u] = 1;
    for (int i = head[u];i != -1;i = edge[i].next)
    {
        v = edge[i].to;
        if (!dfn[v])
        {
            tarjan(v);
            low[u] = low[u] > low[v]?low[v]:low[u];
        }
        else if (vis[v]&&low[u] > dfn[v])
            low[u] = dfn[v];
    }
    if (dfn[u] == low[u])
    {
        ++scc;
        do
        {
            v = stack[--top];
            vis[v] = 0;
            belong[v] = scc;
        } while (u != v);
    }
}
void sovle ()
{
    int u;
    scc = top = cnt = 0;
    memset (dfn,0,sizeof(dfn));
    memset (vis,0,sizeof(vis));
    for (u = 1;u <= n;u ++)
        if (!dfn[u])
            tarjan(u);
}

int topo ()
{
    int u,v,i,cur,count = 0;;
    for (u = 1;u <= scc;u ++)
        if (!indeg[u])
        {
            count ++;
            cur = u;
        }
    if (count > 1)
        return 0;
    int num = scc;
    while (num --)
    {
        count = 0;
        for (i = head1[cur];i != -1;i = edge1[i].next)
        {
            u = edge1[i].frm;
            v = edge1[i].to;
            //indeg[u] --;
            indeg[v] --;
            if (!indeg[v])
            {
                count ++;
                cur = v;
            }
        }
        if (count > 1)
            return 0;
    }
    return 1;
}

int ans()
{
    int u,v,i,count = 0;
    memset (indeg,0,sizeof(indeg));
    memset (map,0,sizeof (map));
    for (u = 1;u <= n;u ++)
        for (i = head[u];i != -1;i = edge[i].next)
        {
            v = edge[i].to;
            int j,k;
            j = belong[u];
            k = belong[v];
            if (j != k&& !map[j][k])
            {
                indeg[k] ++;
                addedge1 (j,k);
                map[j][k] = 1;
            }
        }
    if (topo())
        return 1;
    return 0;

}

int main ()
{
    int t,m,u,v;
    scanf ("%d",&t);
    while (t --)
    {
        p = 0;
        memset (head1,-1,sizeof(head1));
        memset (head,-1,sizeof(head));
        scanf ("%d%d",&n,&m);
        while (m --)
        {
            scanf ("%d%d",&u,&v);
            addedge (u,v);
        }
        sovle ();
        if (ans())
            printf ("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}