标签:tac hdu puts name ++ 1.0 map 产生 memset
题意:给定n个人的两两比赛,每个人要么是good 要么是bad,现在问你能不能唯一确定并且是合理的。
析:其实就是一个二分图染色,如果产生矛盾了就是不能,否则就是可以的。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } vector<int> G[maxn]; set<int> sets; int a[maxn], b[maxn]; int color[maxn]; bool ok; void dfs(int u, int x){ if(!ok) return ; for(int i = 0; i < G[u].size(); ++i){ int v = G[u][i]; if(color[v] && color[v] + color[u] != 3){ ok = false; return ; } if(color[v]) continue; color[v] = x; dfs(v, 3 - x); } } int main(){ int x, y; while(scanf("%d %d %d %d", &n, &m, &x, &y) == 4){ for(int i = 1; i <= n; ++i) G[i].clear(); sets.clear(); for(int i = 0; i < m; ++i){ int u, v; scanf("%d %d", &u, &v); G[u].push_back(v); G[v].push_back(u); sets.insert(u); sets.insert(v); } memset(color, 0, sizeof color); for(int i = 0; i < x; ++i){ scanf("%d", a+i); sets.insert(a[i]); } for(int i = 0; i < y; ++i){ scanf("%d", b+i); sets.insert(b[i]); } if(sets.size() != n){ puts("NO"); continue; } ok = true; for(int i = 0; i < x && ok; ++i){ if(color[a[i]] && color[a[i]] != 1) ok = false; color[a[i]] = 1; dfs(a[i], 2); } for(int i = 0; i < y && ok; ++i){ if(color[b[i]] && color[b[i]] != 2) ok = false; color[b[i]] = 2; dfs(b[i], 1); } for(int i = 1; i <= n && ok; ++i) if(!color[i]){ color[i] = 1; dfs(i, 2); } printf("%s\n", ok ? "YES" : "NO"); } return 0; }
HDU 5971 Wrestling Match (二分图)
标签:tac hdu puts name ++ 1.0 map 产生 memset
原文地址:http://www.cnblogs.com/dwtfukgv/p/6792855.html