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  • HDU 5971 Wrestling Match (二分图)

    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
    ", ok ? "YES" : "NO");
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6792855.html
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