BZOJ1823:[JSOI2010]满汉全席——题解

 

https://www.lydsy.com/JudgeOnline/problem.php?id=1823

https://www.luogu.org/problemnew/show/P4171

题面太长啦就不粘过来啦!

裸的2-SAT用来练板子的。

显然属于“a和b之间必须选一种”模型,只要a'向b连边,b'向a连边即可。

(被这题读入坑了orz)

#include<cstdio>
#include<queue>
#include<cctype>
#include<cstring>
#include<cmath>
#include<stack>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=1005;
const int M=20005;
struct node{
    int to,nxt;
}e[M];
int T,n,m,head[N],cnt;
int dfn[N],low[N],t,l;
int to[N];
bool inq[N];
stack<int>q;
inline int neg(int x){
    if(x>n)return x-n;
    return x+n;
}
inline void add(int u,int v){
    e[++cnt].to=v;e[cnt].nxt=head[u];head[u]=cnt;
}
void tarjan(int u){
    dfn[u]=low[u]=++t;
    q.push(u);inq[u]=1;
    for(int i=head[u];i;i=e[i].nxt){
        int v=e[i].to;
        if(!dfn[v]){
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }else if(inq[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u]){
        int v;l++;
        do{
            v=q.top();q.pop();
            inq[v]=0;to[v]=l;
        }while(u!=v);
    }
}
inline void init(){
    cnt=t=l=0;
    memset(head,0,sizeof(head));
    memset(dfn,0,sizeof(dfn));
}
inline int get(){
    int k;char ch=0;
    while(ch!='m'&&ch!='h')ch=getchar();
    scanf("%d",&k);
    if(ch=='h')k+=n;
    return k;
}
int main(){
    scanf("%d",&T);
    while(T--){
        init();
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++){
            int a=get(),b=get();
            add(neg(a),b);
            add(neg(b),a);
        }
        for(int i=1;i<=2*n;i++)
            if(!dfn[i])tarjan(i);
        bool flag=1;
        for(int i=1;i<=n&&flag;i++){
            if(to[i]==to[i+n])flag=0;
        }
        if(flag)puts("GOOD");
        else puts("BAD");
    }
}

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 +本文作者:luyouqi233。               +

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posted @ 2018-05-10 22:17  luyouqi233  阅读(159)  评论(0编辑  收藏  举报