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
;
int pre
, lowlink
, sccno
, dfs_clock, scc_cnt;
int n, m, top;
int S
;
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");
}
}