HDU5971 Wrestling Match（二分图的判断）

Wrestling Match

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1027    Accepted Submission(s): 388

Problem Description

Nowadays, at least one wrestling match is held every year in our country. There are a lot of people in the game is "good player”, the rest is "bad player”. Now, Xiao Ming is referee of the wrestling match and he has a list of the matches in his hand. At the same time, he knows some people are good players,some are bad players. He believes that every game is a battle between the good and the bad player. Now he wants to know whether all the people can be divided into "good player" and "bad player".

 

Input

Input contains multiple sets of data.For each set of data,there are four numbers in the first line:N (1 ≤ N≤ 1000)、M(1 ≤M ≤ 10000)、X,Y(X+Y≤N ),in order to show the number of players(numbered 1toN ),the number of matches,the number of known "good players" and the number of known "bad players".In the next M lines,Each line has two numbersa, b(a≠b) ,said there is a game between a and b .The next line has X different numbers.Each number is known as a "good player" number.The last line contains Y different numbers.Each number represents a known "bad player" number.Data guarantees there will not be a player number is a good player and also a bad player.

 

Output

If all the people can be divided into "good players" and "bad players”, output "YES", otherwise output "NO".

 

Sample Input

5 4 0 01 31 43 54 55 4 1 01 31 43 54 52

 

Sample Output

NOYES

 

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

思路：这个题目表达的意思有点模糊，从事例上来说，题目让求得是能否把给定的顶点完全分成两个集合，即一部分分成good，一部分分成bad，这样就是判断给定的边能否成为一个二分图。

#include <cstdio>#include <vector>#include <cstring>using namespace std;int n,m,x,y,color[1005],vis[1005];vector<int>p[1005];int dfs(int u){for(int i = 0; i < p[u].size(); i ++){//对以该顶点为头的链表进行判断 int v = p[u][i];//取其中的节点 if(!color[v]){//当前点是未知点 color[v] = !color[u];//为头节点的相反状态 if(!dfs(v))//判断是否存在与v节点相矛盾的顶点 return 0;}else if(color[v] == color[u]){//两顶点的状态一致 return 0;}}return 1;}int main(){while(~scanf("%d%d%d%d",&n,&m,&x,&y)){for(int i = 0; i <= 1000; i ++)//对数组进行清零 p[i].clear();memset(color,0,sizeof(color));memset(vis,0,sizeof(vis));int a,b;for(int i = 0; i < m; i ++){scanf("%d%d",&a,&b);p[a].push_back(b);//把b放在a的后面 vis[a] = vis[b] = 1;//对出现的元素进行标记 }for(int i = 0; i < x; i ++){scanf("%d",&a);color[a] = vis[a] = 1;//对确定是good的顶点进行标记 }for(int i = 0; i < y; i ++){//对于bad的顶点可以作为位置顶点来判断，因为就两个集合 scanf("%d",&a);vis[a] = 1;}int cnt = 0;for(int i = 1; i <= n; i ++){//计算出新的顶点数目 if(vis[i])cnt ++;}if(n == 1 || cnt < n){//顶点数目为1或者出现的顶点数小于n，则输出NO printf("NO\n");continue;}int flag = 1;for(int i = 1; i <= n; i ++){if(!dfs(i)){//判断是否存在与其相矛盾的顶点 flag = 0;break;}}if(flag)printf("YES\n");elseprintf("NO\n");}return 0;}