Wrestling Match
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2539 Accepted Submission(s): 922
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 0
1 3
1 4
3 5
4 5
5 4 1 0
1 3
1 4
3 5
4 5
2
Sample Output
NO
YES
虽说是道水题但是能够一遍过还是挺爽的。
题意：有n个选手，m场比赛，x个著名好选手，y个著名坏选手，每场比赛都是一名号选手与一名坏选手对打。问在这些条件下是否让所有的选手都拥有身份。（非好即坏
解题思路：当时想的就是搜索，先把与x相关的搜一遍，确认关系，再把与y相关的搜一遍，最后剩下都不相关的看是否有比赛，如果有比赛，随机设1名为好选手，搜索和他相关的选手，直到所有的比赛都遍历完。
ac代码：
1 #include <cstdio> 2 #include <iostream> 3 #include <cmath> 4 #include <cstring> 5 #include <algorithm> 6 #define ll long long 7 const int maxn = 1000+10; 8 using namespace std; 9 int mp[maxn][maxn]; 10 int vis[maxn]; 11 int n,m,x,y; 12 13 int ed=0; //ed==1的时候，说明该选手没有比赛，ed==2的时候，说明遇到两个同类型的选手在一起比赛，直接f=1，输出NO 14 void df(int a,int s) 15 { 16 int i=1; 17 if(ed) return; 18 for(i=1;i<=n;++i) 19 { 20 if(mp[a][i] && vis[i]!=s && vis[i]!=0) 21 { 22 ed=2; 23 // printf(" %d %d ",a,i); 24 break; 25 } 26 if(mp[a][i] && !vis[i]) 27 { 28 vis[i]=s; 29 if(s==1) 30 df(i,2); 31 else 32 df(i,1); 33 } 34 } 35 if(i==n+1) 36 { 37 ed=1; 38 return; 39 } 40 } 41 int main() { 42 43 while(~scanf("%d%d%d%d",&n,&m,&x,&y)) 44 { 45 memset(mp,0,sizeof(mp)); 46 memset(vis,0,sizeof(vis)); 47 int a,b; 48 for(int i=0;i<m;++i) 49 { 50 scanf("%d%d",&a,&b); 51 mp[a][b]=1; 52 mp[b][a]=1; 53 } 54 55 int f=0; 56 57 for(int i=0;i<x;++i) 58 { 59 scanf("%d",&a); 60 vis[a]=1; 61 ed=0; 62 df(a,2); 63 } 64 65 if(ed==2) f=1; 66 67 for(int i=0;i<y;++i) 68 { 69 scanf("%d",&b); 70 vis[b]=2; 71 ed=0; 72 df(b,1); 73 } 74 75 if(ed==2) f=1; 76 77 for(int i=1;i<=n;++i) 78 { 79 for(int j=1;j<=n;++j) 80 { 81 ed=0; 82 if(mp[i][j] && vis[i]==0 && vis[j]==0) 83 { 84 vis[i]=1; 85 df(i,2); 86 } 87 } 88 } 89 90 for(int i=1;i<=n;++i) 91 { 92 if(vis[i]==0) 93 { 94 f=1; 95 break; 96 } 97 } 98 if(!f) 99 printf("YES "); 100 else 101 printf("NO "); 102 } 103 return 0; 104 }