poj 3207 2-sat

思路：把每一条线段当做点，则有“内”、“外”两种取值，然后在冲突的线段之间连边，判断是否有解即可。

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <cstring>
  4 #include <cstdio>
  5 using namespace std;
  6 
  7 const int N = 500;
  8 const int M = 1000000;
  9 int head[N * 2];
 10 int s[N * 2];
 11 bool mark[N * 2];
 12 int n, e, c;
 13 
 14 struct Edge
 15 {
 16     int v, next;
 17 } edge[M];
 18 
 19 void addEdge( int u, int v )
 20 {
 21     edge[e].v = v;
 22     edge[e].next = head[u];
 23     head[u] = e++;
 24 }
 25 
 26 void init()
 27 {
 28     e = 0;
 29     memset( head, -1, sizeof(head) );
 30     memset( mark, false, sizeof(mark) );
 31 }
 32 
 33 bool dfs( int u )
 34 {
 35     if ( mark[u ^ 1] ) return false;
 36     if ( mark[u] ) return true;
 37     mark[u] = true;
 38     s[c++] = u;
 39     for ( int i = head[u]; i != -1; i = edge[i].next )
 40     {
 41         int v = edge[i].v;
 42         if ( !dfs(v) ) return false;
 43     }
 44     return true;
 45 }
 46 
 47 bool solve()
 48 {
 49     for ( int i = 0; i < 2 * n; i += 2 )
 50     {
 51         if ( !mark[i] && !mark[i + 1] )
 52         {
 53             c = 0;
 54             if ( !dfs(i) )
 55             {
 56                 while ( c )
 57                 {
 58                     c--;
 59                     mark[s[c]] = false;
 60                 }
 61                 if ( !dfs( i + 1 ) ) return false;
 62             }
 63         }
 64     }
 65     return true;
 66 }
 67 
 68 struct T 
 69 {
 70     int x, y;
 71 } t[N];
 72 
 73 int main ()
 74 {
 75     scanf("%d%d", &n, &n);
 76     init();
 77     for ( int i = 0; i < n; i++ )
 78     {
 79         scanf("%d%d", &t[i].x, &t[i].y);
 80         if ( t[i].x > t[i].y ) swap( t[i].x, t[i].y );
 81         for ( int j = 0; j < i; j++ )
 82         {
 83             if ( ( t[i].x < t[j].x && t[i].y < t[j].y && t[i].y > t[j].x ) 
 84               || ( t[i].x > t[j].x && t[i].y > t[j].y && t[i].x < t[j].y ) )
 85             {
 86                 int li = i << 1, ri = li | 1;
 87                 int lj = j << 1, rj = lj | 1;
 88                 addEdge( li, rj );
 89                 addEdge( ri, lj );
 90                 addEdge( lj, ri );
 91                 addEdge( rj, li );
 92             }
 93         }
 94     }
 95     if ( solve() )
 96     {
 97         puts("panda is telling the truth...");
 98     }
 99     else
100     {
101         puts("the evil panda is lying again");
102     }
103     return 0;
104 }

 

posted @ 2015-08-21 15:03  hxy_has_been_used  阅读(127)  评论(0)  编辑  收藏  举报