Going from u to v or from v to u?
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15755 | Accepted: 4172 |
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
题意:是给出一些点，给出n个点和m条边,接着给出直接相连的边(注意是有向边),求解随意x,y两点间是否存在 x 可到达 y 或者y可
到达x,假设随意x和y都满足这种条件,就输出"Yes", 否则输出"No".。
注意。这里是 x 到达 y 或者 y 到达 x ，是或者不是并且 。！
。
假设是“并且”的话。非常明显的是推断整个图是否为一个强连通分量（如题HDU 1296 , 题目解析）。但这题并非这样。
本题应推断整个图是否为一个弱连通分量。
正确思路：先求解出该有向图的强连通分量。然后依据求解出来的强连通分量进行缩点又一次建图
问题转换为求解在新图中是否存在一条能走全然部的顶点的路径，这时能够对缩点后的新图进行拓扑排序，看拓扑排序能否够成功进行。
拓扑排序遵循条件
一：新图不能有多于1个的入度为0的点，这是保证每一个点都有边相连。
二：在拓扑排序遍历点u的过程中，若去掉与u相关的边后出现多于1个的入度为0的点，说明这些点仅仅能由u到达，而它们之间不存在可达路径。这时不满足弱连通，跳出。
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#define maxn 10000 + 100
#define maxm 100000 + 1000
using namespace std;
int n, m;
struct node {
int u, v, next;
};
node edge[maxm];
int head[maxn], cnt;
int low[maxn], dfn[maxn];
int dfs_clock;
int Stack[maxn];
bool Instack[maxn];
int top;
int Belong[maxn] , scc_clock;
int in[maxn];
vector<int>Map[maxn];
void init(){
cnt = 0;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v){
edge[cnt] = {u, v, head[u]};
head[u] = cnt++;
}
void getmap(){
scanf("%d%d", &n, &m);
while(m--){
int a, b;
scanf("%d%d", &a, &b);
addedge(a, b);
}
}
void tarjan(int u, int per){
int v;
low[u] = dfn[u] = ++dfs_clock;
Stack[top++] = u;
Instack[u] = true;
for(int i = head[u]; i != -1; i = edge[i].next){
v = edge[i].v;
if(!dfn[v]){
tarjan(v, u);
low[u] = min(low[v], low[u]);
}
else if(Instack[v]){
low[u] = min(low[u], dfn[v]);
}
}
if(dfn[u] == low[u]){
scc_clock++;
do{
v = Stack[--top];
Instack[v] = false;
Belong[v] = scc_clock;
}while(u != v);
}
}
void find(){
memset(low, 0, sizeof(low));
memset(dfn, 0, sizeof(dfn));
memset(Instack, false, sizeof(Instack));
memset(Belong, 0, sizeof(Belong));
dfs_clock = scc_clock = top = 0;
for(int i = 1; i <= n; ++i){
if(!dfn[i])
tarjan(i, i);
}
}
void suodian(){
for(int i = 1; i <= scc_clock; ++i){
Map[i].clear();
in[i] = 0;
}
for(int i = 0; i < cnt; ++i){
int u = Belong[edge[i].u];
int v = Belong[edge[i].v];
if(u != v){
Map[u].push_back(v);
in[v]++;
}
}
}
void solve(){
queue<int>q;
int num = 0;
for(int i = 1; i <= scc_clock; ++i){
if(!in[i]){
num++;
q.push(i);
}
if(num > 1){
printf("No\n");
return ;
}
}
while(!q.empty()){
int u = q.front();
q.pop();
num = 0;
for(int i = 0; i < Map[u].size(); ++i){
int v = Map[u][i];
in[v]--;
if(!in[v]){
num++;
//有两个或两个以上的分支。不是弱连通
if(num > 1){
printf("No\n");
return ;
}
q.push(v);
}
}
}
printf("Yes\n");
}
int main (){
int T;
scanf("%d", &T);
while(T--){
init();
getmap();
find();
suodian();
solve();
}
return 0;
}