POJ 3678 Kath Puzzle （2-sat）

首先要说，写代码，要仔细；

其次要说，在poj，当保证了自己代码和算法几乎没错误，思路也正确的时候，还超时，检查一下你提交的编译类型是不是G++或者是GCC，如果是，改成C++试试！

这道题很典型的2-sat!

关于这个道题，网上已经有很多解释了，在这里不再赘述！

代码如下：（个人认为，tarjan比那个konaba……的算法更简单一些）

#include <cstdio>#include <algorithm>#include <iostream>#include <cstring>#include <string>#include <vector>#include <stack>using namespace std;const int N = 2020;vector <int> G[N];int pre[N], lowlink[N], sccno[N], dfs_clock, scc_cnt;int n, m, top;int S[N];string op;void addedge( int i, int j ){    G[i].push_back(j);}void Tarjan( int u ) {    pre[u] = lowlink[u] = ++dfs_clock;    S[top++] = u;    for ( int i = 0; i < G[u].size(); ++i ) {        int v = G[u][i];        if ( !pre[v] ) {            Tarjan(v);            lowlink[u] = min( lowlink[u], lowlink[v] );        }        else if ( !sccno[v] ) {            lowlink[u] = min( pre[v], lowlink[u] );        }    }    if ( pre[u] == lowlink[u] ) {        scc_cnt++;        while ( 1 ) {            int x = S[--top];             sccno[x] = scc_cnt;            if ( x == u ) break;        }    }}void find_scc( int nodenum ){    dfs_clock = scc_cnt = top = 0;    memset(sccno, 0, sizeof(sccno));    memset(pre, 0, sizeof(pre));    for ( int i = 0; i < nodenum; ++i )          if ( !pre[i] ) Tarjan(i);}int main(){    while ( scanf("%d%d", &n, &m) != EOF ) {        for ( int i = 0; i <= n*2; ++i ) G[i].clear();        while ( m-- ) {            int u, v, cc;            cin >> u >> v >> cc >> op;            if ( op[0] == 'A' ) {                if ( cc == 1 ) {                    addedge( 2*u+1, 2*u );                    addedge( 2*v+1, 2*v );                    //addedge( 2*v, 2*u );                    //addedge( 2*u, 2*v );                }                else {                    addedge( 2*u, 2*v+1 );                    addedge( 2*v, 2*u+1 );                }            }            else if ( op[0] == 'O' ) {                if ( cc == 1 ) {                    addedge( 2*u+1, 2*v );                    addedge( 2*v+1, 2*u );                }                else {                    addedge( 2*u, 2*u+1 );                    addedge( 2*v, 2*v+1 );                    //addedge( 2*v+1, 2*u+1 );                    //addedge( 2*u+1, 2*v+1 );                }            }            else if ( op[0] == 'X' ) {                if ( cc == 1 ) {                    addedge( 2*u, 2*v+1 );                    addedge( 2*v, 2*u+1 );                    addedge( 2*u+1, 2*v );                    addedge( 2*v+1, 2*u );                }                else {                    addedge( 2*u, 2*v );                    addedge( 2*v, 2*u );                    addedge( 2*u+1, 2*v+1 );                    addedge( 2*v+1, 2*u+1 );                }            }        }        find_scc( n*2 );        bool is = true;        for ( int i = 0; i < n; i++ )            if ( sccno[i*2] == sccno[i*2+1] ) {                is = false;                break;            }        if ( is ) printf("YES\n");        else printf("NO\n");    }}