题意：
有n个候选人和m张选票，每张选票上有一种限制，问是否有一种让m张选票上的限制都满足的选举方案。
分析：
2-sat判断是否有解，不需输出解，标准化操作。
代码：
//poj 3905
//sep9
#include <iostream>
#include <stack>
using namespace std;
const int maxN=2024;
const int maxM=2200000;
int e,n,m,t,ecnt;
int head[maxN],ins[maxN],low[maxN],dfn[maxN];
int sol[maxN],belong[maxN];
stack<int> s;
struct Edge
{int v,next;
}edge[maxM];void addegde(int u,int v)
{edge[e].v=v;edge[e].next=head[u];head[u]=e++;
}void dfs(int x)
{low[x]=dfn[x]=++t; s.push(x);ins[x]=1;for(int i=head[x];i!=-1;i=edge[i].next){int v=edge[i].v;if(!dfn[v]){dfs(v);low[x]=min(low[x],low[v]);}else if(ins[v]==1)low[x]=min(low[x],dfn[v]);} if(dfn[x]==low[x]){++ecnt; int k;do{k=s.top();s.pop();ins[k]=0;belong[k]=ecnt;}while(dfn[k]!=low[k]);}
}
bool two_sat()
{memset(ins,0,sizeof(ins)); memset(dfn,0,sizeof(dfn)); while(!s.empty()) s.pop();int i;t=0,ecnt=0;for(i=1;i<=2*n;++i)if(!dfn[i])dfs(i);for(i=1;i<=n;++i)if(belong[i]==belong[i+n])return false;return true;
}int deal()
{int a,b,c;char op1[8],op2[8];e=0;memset(head,-1,sizeof(head));while(m--){scanf("%s%s",op1,op2);sscanf(op1+1,"%d",&a);sscanf(op2+1,"%d",&b); if(op1[0]=='+'&&op2[0]=='+'){addegde(a+n,b);addegde(b+n,a);}else if(op1[0]=='-'&&op2[0]=='-'){addegde(a,b+n);addegde(b,a+n);}else if(op1[0]=='+'&&op2[0]=='-'){addegde(a+n,b+n);addegde(b,a);}else if(op1[0]=='-'&&op2[0]=='+'){addegde(a,b);addegde(b+n,a+n);} }bool ans=two_sat();printf("%d\n",ans==true?1:0);return 0;
} int main()
{while(scanf("%d%d",&n,&m)==2){deal();}return 0;
}