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

u到v或者v到u

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

简单来说,这是一个有向图的弱连通问题…

注意，是有向图！

学习了图的强连通后，不难发现，一个样有向图的强连通是有向环，对于环上的每一个点(room),他们都可以互相到达。

如果我们找到图中的强连通分量，将他们缩点，再连边，会形成一个“压缩到极致的”无环有向图(不想无向图，缩点连边后是一棵树。实际上是因为有向图有横叉边)

再仔细想想，如果此时的新图中，有两个或以上入度为0的点，那么这两个点肯定不能互相到达。

如果只有一个入度为0的点，以这个点为root，DFS遍历整个图，如果对于某个结点，他有两个或以上的儿子，那么这两棵或以上的子树上的点，就不能互相到达。说了这么多其实就是能形成一条单链。。。

那么，就来实现一下。

代码：

#include<cstdio>#include<stack>#include<cstring>#include<algorithm>using namespace std;#define MAXN 1000#define MAXM 6000int t,n,m;struct node{    int v;    node *next;}edge[MAXM+10],*adj[MAXN+10],*ecnt=&edge[0];int dfn[MAXN+10],low[MAXN+10],dcnt,block_cnt,bccno[MAXN+10],in[MAXN+10],out[MAXN+10];stack<int> point;bool instack[MAXN+10],mat[MAXN+10][MAXN+10],vis[MAXN+10];void reset(){    memset(edge,0,sizeof(edge)); memset(adj,0,sizeof(adj)); ecnt=&edge[0];    memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(low)); memset(bccno,0,sizeof(bccno));    memset(in,0,sizeof(in)); memset(out,0,sizeof(out));    memset(instack,0,sizeof(instack)); memset(mat,0,sizeof(mat)); memset(vis,0,sizeof(vis));    dcnt=block_cnt=0;}void addedge(int u,int v){    node *p=++ecnt;    p->v=v;    p->next=adj[u];    adj[u]=p;}void read(){    int x,y;    scanf("%d%d",&n,&m);    for(int i=1;i<=m;i++){        scanf("%d%d",&x,&y);        addedge(x,y);        mat[x][y]=true;    }}void dfs_find_SCC(int u,int fa){    point.push(u);    instack[u]=true;    dfn[u]=low[u]=++dcnt;    for(node *p=adj[u];p;p=p->next){        int v=p->v;        if(!dfn[v]){            dfs_find_SCC(v,u);            low[u]=min(low[u],low[v]);        }        else if(instack[v])            low[u]=min(low[u],dfn[v]);    }    if(dfn[u]==low[u]){        ++block_cnt;        int uu;        do{            uu=point.top(); point.pop();            bccno[uu]=block_cnt;            instack[uu]=false;        }while(uu!=u);    }}void build_new_tree() {    memset(edge,0,sizeof(edge));    memset(adj,0,sizeof(adj));    ecnt=&edge[0];    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++){            if(j==i) continue;            if(mat[i][j]&&bccno[i]!=bccno[j]){                addedge(bccno[i],bccno[j]);                out[bccno[i]]++,in[bccno[j]]++;            }        }}bool dfs_Judge(int u){    vis[u]=true;    int cnt=0;    for(node *p=adj[u];p;p=p->next){        int v=p->v;        if(!vis[v]){            cnt++;            bool f=dfs_Judge(v);            if(!f) return false;        }    }    if(cnt>=2) return false;    return true;}bool solve(){    //最多只能有一个入度为0的点，如果有多个入度为0的点，则这两个连通分量肯定是不连通的    int incnt=0,rt=-1;    for(int i=1;i<=block_cnt;i++)        if(!in[i]) incnt++,rt=i;    if(incnt>1) return false;    if(dfs_Judge(rt)) return true;    else return false;}int main(){    scanf("%d",&t);    for(int i=1;i<=t;i++){        reset();        read();        for(int i=1;i<=n;i++)            if(!dfn[i])                dfs_find_SCC(i,-1); //strongly connected componenet:强连通        build_new_tree();  //缩点，建树        bool f=solve();        if(f) printf("Yes\n");        else printf("No\n");    }}

不大满意的地方:reset写得比较丑。。。。