http://acm.hdu.edu.cn/showproblem.php?pid=5971
Wrestling Match
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 25 Accepted Submission(s): 15
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
Source
Recommend
給定一個圖，有可能是分散的圖，其中有一些點是固定是顏色的，現在要求判斷其能否成為二分圖。
假如是分成了若干個聯通快（塊內的點個數 >= 2），對於每個聯通快，如果有一些點是確定了的，那么就應該選那個點進行開始染色，途中如果遇到一些點已經確定顏色的了，但是和現在的想填的顏色不同，那么就應該輸出NO，否則，進行染色即可。
對於點數為1的聯通快，如果它沒有被確定顏色的話，那么就直接輸出NO了。
然后邊數要開兩倍，不然直接給wa，這里坑了我。一直做不出。
5 3 0 0 1 2 1 3 4 5
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> int n, m, x, y; const int maxn = 1000 + 20; struct node { int u, v, w; int tonext; } e[2 * 10000 + 20]; int first[maxn]; bool vis[maxn]; int black = 1; int white = 0; int arr[maxn]; int ca[maxn]; bool in[maxn]; bool flag; int num; void add(int u, int v, int w) { ++num; e[num].u = u; e[num].v = v; e[num].w = w; e[num].tonext = first[u]; first[u] = num; } void dfs(int cur, int col) { for (int i = first[cur]; i && flag; i = e[i].tonext) { int v = e[i].v; if (vis[v]) { if (arr[v] == col) { flag = false; return; } } if (vis[v]) continue; vis[v] = true; if (arr[v] == -1) { arr[v] = !col; dfs(v, !col); } else { if (arr[v] == col) { flag = false; return; } else { dfs(v, !col); } } } } void work() { num = 0; memset(arr, -1, sizeof arr); memset(ca, -1, sizeof ca); memset(in, 0, sizeof in); memset(first, 0, sizeof first); flag = true; for (int i = 1; i <= m; ++i) { int u, v; cin >> u >> v; add(u, v, 1); add(v, u, 1); in[v] = in[u] = 1; } for (int i = 1; i <= x; ++i) { int val; cin >> val; ca[val] = black; arr[val] = black; } for (int i = 1; i <= y; ++i) { int val; cin >> val; ca[val] = white; arr[val] = white; } for (int i = 1; i <= n; ++i) { if (in[i] == 0 && ca[i] == -1) { cout << "NO" << endl; return; } } memset(vis, 0, sizeof vis); for (int i = 1; i <= n; ++i) { if (vis[i]) continue; if (ca[i] == -1) continue; vis[i] = 1; arr[i] = ca[i]; dfs(i, ca[i]); } for (int i = 1; i <= n; ++i) { if (vis[i]) continue; arr[i] = black; dfs(i, black); } if (flag == false) { cout << "NO" << endl; return; } cout << "YES" << endl; } int main() { #ifdef local freopen("data.txt","r",stdin); #endif IOS; while (cin >> n >> m >> x >> y) { work(); } return 0; }