cf 167.div2 E.Dima and Horses
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题意很简单,就是分组,让冲突最多有一种,因为只要分成2组,就是个2-SAT
但是最多只有3个不和,就是不和的三个里至少有2个在一组,所以应该必然有解(这边不太明白,感觉这样),直接染色就可以了
/*author:jxylang:C/C++university:China,Xidian University**If you need to reprint,please indicate the source***/#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <queue>#define INF 1E9using namespace std;int en[300010][3];int d[300010];bool ok[300010];bool sat(int v){ int i,now=0,u; for(i=0;i<d[v];i++) { u=en[v][i]; if(ok[u]==ok[v])now++; } if(now>1) { ok[v]=!ok[v]; for(i=0;i<d[v];i++) { u=en[v][i]; if(ok[v]==ok[u])sat(u); } }}int main(){ int n,m,i,a,b; scanf("%d%d",&n,&m); memset(d,0,sizeof(d)); memset(ok,0,sizeof(ok)); for(i=0;i<m;i++) { scanf("%d%d",&a,&b); en[a][d[a]++]=b; en[b][d[b]++]=a; } for(i=1;i<=n;i++) sat(i); for(i=1;i<=n;i++) printf("%d",ok[i]); puts("");}
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