【POJ】3678 Katu Puzzle

简单2-SAT。

  1 /* 3678 */
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <cstdio>
  5 #include <cstring>
  6 #include <cstdlib>
  7 using namespace std;
  8 
  9 #define MAXN 1005
 10 #define MAXE 1000005
 11 
 12 typedef struct {
 13     int v, next;
 14 } Edge_t;
 15 
 16 int pre[MAXN*2], low[MAXN*2];
 17 int bn[MAXN*2];
 18 int dfs_clock, block;
 19 int S[MAXN*2], top;
 20 int head[MAXN*2], L;
 21 Edge_t E[MAXE];
 22 int n, m, n2;
 23 int a[MAXE], b[MAXE], c[MAXE];
 24 char op[MAXE];
 25 
 26 void addEdge(int u, int v) {
 27     E[L].v = v;
 28     E[L].next = head[u];
 29     head[u] = L++;
 30 }
 31 
 32 void init() {
 33     dfs_clock = block = top = L = 0;
 34     memset(pre, 0, sizeof(int)*n2);
 35     memset(bn, 0, sizeof(int)*n2);
 36     memset(head, -1, sizeof(int)*n2);
 37 }
 38 
 39 void dfs(int u) {
 40     int i, j, k, v;
 41     
 42     S[top++] = u;
 43     pre[u] = low[u] = ++dfs_clock;
 44     for (i=head[u]; i!=-1; i=E[i].next) {
 45         v = E[i].v;
 46         if (!pre[v]) {
 47             dfs(v);
 48             low[u] = min(low[u], low[v]);
 49         } else if (!bn[v]) {
 50             low[u] = min(low[u], pre[v]);
 51         }
 52     }
 53     if (low[u] == pre[u]) {
 54         ++block;
 55         do {
 56             bn[S[--top]] = block;
 57         } while (S[top]!=u);
 58     }
 59 }
 60 
 61 
 62 
 63 int main() {
 64     int i, j, k;
 65     int ii, jj;
 66     char cmd[8];
 67     bool flag;
 68     
 69     #ifndef ONLINE_JUDGE
 70         freopen("data.in", "r", stdin);
 71         freopen("data.out", "w", stdout);
 72     #endif
 73     
 74     while (scanf("%d %d",&n,&m)!=EOF) {
 75         n2 = n+n;
 76         for (i=0; i<m; ++i) {
 77             scanf("%d %d %d %s", &a[i], &b[i], &c[i], cmd);
 78             op[i] = cmd[0];
 79         }
 80         flag = true;
 81         for (k=0; flag&&k<32; ++k) {
 82             init();
 83             for (i=0; i<m; ++i) {
 84                 ii = a[i]+a[i];
 85                 jj = b[i]+b[i];
 86                 if (op[i] == 'A') {
 87                     // and
 88                     if (c[i] & (1<<k)) {
 89                         addEdge(ii, ii+1);
 90                         addEdge(ii+1, jj+1);
 91                         addEdge(jj, jj+1);
 92                         addEdge(jj+1, ii+1);
 93                     } else {
 94                         addEdge(ii+1, jj);
 95                         addEdge(jj+1, ii);
 96                     }
 97                 } else if (op[i] == 'O') {
 98                     // or
 99                     if (c[i] & (1<<k)) {
100                         addEdge(ii, jj+1);
101                         addEdge(jj, ii+1);
102                     } else {
103                         addEdge(ii, jj);
104                         addEdge(jj, ii);
105                         addEdge(ii+1, ii);
106                         addEdge(jj+1, jj);
107                     }
108                 } else if (op[i] == 'X') {
109                     // xor
110                     if (c[i] & (1<<k)) {
111                         addEdge(ii, jj+1);
112                         addEdge(ii+1, jj);
113                         addEdge(jj, ii+1);
114                         addEdge(jj+1, ii);
115                     } else {
116                         addEdge(ii, jj);
117                         addEdge(ii+1, jj+1);
118                         addEdge(jj, ii);
119                         addEdge(jj+1, ii+1);
120                     }
121                 }
122             }
123             for (i=0; i<n2; ++i)
124                 if (!pre[i])
125                     dfs(i);
126             for (i=0; flag&&i<n2; i+=2)
127                 if (bn[i] == bn[i+1])
128                     flag = false;
129         }
130         if (flag)
131             puts("YES");
132         else
133             puts("NO");
134     }
135     
136     return 0;
137 }

 

posted on 2015-03-16 22:10  Bombe  阅读(277)  评论(0)  编辑  收藏  举报

导航