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  • UVa 11294 Wedding (TwoSat)

    题意:有 n-1 对夫妻参加一个婚宴,所有人都坐在一个长长的餐桌左侧或者右侧,新郎和新娘面做面坐在桌子的两侧。由于新娘的头饰很复杂,她无法看到和她坐在同一侧餐桌的人,只能看到对面餐桌的人。任意一对夫妻不能坐在桌子的同侧,另外有m对人吵过架,而新娘不希望看到两个吵过架的人坐在他的对面,问如何安排这些座位。

    析:很明显的TwoSat问题,假设mark[i<<1]标记了,就是和新娘同侧,否则就是和新娘对侧,这样的话,对于每对吵架的人,他们要么在不同的侧,要么在新娘同侧,注意一定要注意要提前把新娘标记了,而且题目有可能是新娘或者新郎吵过架,RE了好多次,还以为是数组开小了。。。

    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>
    #include <list>
    #include <assert.h>
    #include <bitset>
    #include <numeric>
    #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 fi first
    #define se second
    #define pb push_back
    #define sqr(x) ((x)*(x))
    #define ms(a,b) memset(a, b, sizeof a)
    #define sz size()
    #define pu push_up
    #define pd push_down
    #define cl clear()
    #define all 1,n,1
    #define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
    #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<LL, int> P;
    const int INF = 0x3f3f3f3f;
    const LL LNF = 1e17;
    const double inf = 1e20;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 60 + 100;
    const int maxm = 1e6 + 5;
    const int mod = 10007;
    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;
    }
    
    struct TwoSat{
      int n;
      vector<int> G[maxn<<1];
      bool mark[maxn<<1];
      int S[maxn<<1], c;
    
      void init(int n){
        this-> n = n;
        for(int i = 0; i < (n<<1); ++i)  G[i].cl;
        ms(mark, 0);  mark[0] = 1;
      }
    
      void add_clause(int x, int xval, int y, int yval){
        x = x << 1 | xval;
        y = y << 1 | yval;
        G[x^1].pb(y);
        G[y^1].pb(x);
      }
    
      bool dfs(int x){
        if(mark[x^1])  return false;
        if(mark[x])  return true;
        mark[x] = true;
        S[c++] = x;
        for(int i = 0; i < G[x].sz; ++i)
          if(!dfs(G[x][i]))  return false;
        return true;
      }
    
      bool solve(){
        for(int i = 0; i < (n<<1); i += 2)
          if(!mark[i] && !mark[i^1]){
            c = 0;
            if(!dfs(i)){
              while(c > 0)  mark[S[--c]] = 0;
              if(!dfs(i^1))  return false;
            }
          }
        return true;
      }
    };
    TwoSat twosat;
    
    int main(){
      while(scanf("%d %d", &n, &m) == 2 && n+m){
        twosat.init(n);
        int x, y;
        char c1, c2;
        while(m--){
          scanf("%d%c %d%c", &x, &c1, &y, &c2);
          twosat.add_clause(x, c1 == 'h', y, c2 == 'h');
        }
        if(!twosat.solve()){ puts("bad luck");  continue;  }
        for(int i = 1; i < n; ++i)
          printf("%d%c%c", i, "hw"[twosat.mark[i<<1]], " 
    "[i+1==n]);
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7732372.html
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