标签：tar ota ifd memset sync present tor tee get
http://acm.hdu.edu.cn/showproblem.php?pid=5971
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 25 Accepted Submission(s): 15
给定一个图，有可能是分散的图，其中有一些点是固定是颜色的，现在要求判断其能否成为二分图。
假如是分成了若干个联通快（块内的点个数 >= 2），对于每个联通快，如果有一些点是确定了的，那么就应该选那个点进行开始染色，途中如果遇到一些点已经确定颜色的了，但是和现在的想填的颜色不同，那么就应该输出NO，否则，进行染色即可。
对于点数为1的联通快，如果它没有被确定颜色的话，那么就直接输出NO了。
然后边数要开两倍，不然直接给wa，这里坑了我。一直做不出。
#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; }
hdu 5971 Wrestling Match 判断能否构成二分图
标签：tar ota ifd memset sync present tor tee get
原文地址：http://www.cnblogs.com/liuweimingcprogram/p/6035995.html
