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HomeDatabaseMysql TutorialPOJ 3648 Wedding(2

POJ 3648 Wedding(2-SAT) http://poj.org/problem?id=3648 题意: 有一对新人结婚,n-1对夫妇去参加婚.有一个很长的座子,新娘与新郎坐在座子的两边(相反).接下来n-1对夫妇就坐,其中任何一对夫妇都不能坐在同一边,且(有一些人有奸情)这些有奸情的两个人不能同时

POJ 3648 Wedding(2-SAT)

http://poj.org/problem?id=3648

题意:

        有一对新人结婚,n-1对夫妇去参加婚礼.有一个很长的座子,新娘与新郎坐在座子的两边(相反).接下来n-1对夫妇就坐,其中任何一对夫妇都不能坐在同一边,且(有一些人有奸情)这些有奸情的两个人不能同时坐在新娘对面.(只能分开做,或者都坐到新娘一边去)。对于每个输入实例,输出应该坐在新娘同一边的人编号。

分析:

        由于有n对夫妇(0号表示新婚夫妻).所以我们这里用0表示第0对的妻子,1表示第0对的丈夫. 2*i表示第i对的夫人,2*i+1表示第i对的丈夫.一共就有2*n个人了.

        然后对于每个人来说,把他分成两个节点,如果该人在做左边就mark[i*2],如果该人坐右边就mark[i*2+1].

        我令新娘直接坐左边即第0个人mark[0]=true,新郎直接坐右边即第1个人mark[1*2+1]=true.

        然后对于每对夫妻,因为他们不能在同一边,所以第i对夫妻中a= 2*i表示妻子,b=2*i+1表丈夫. 有这样的关系:

a在左边,那么b就在右边,a*2->b*2+1

a在右边,那么b就在左边,a*2+1->b*2

b在左边,那么a就在右边,b*2->a*2+1

b在右边,那么a就在左边,b*2+1->a*2

        然后对于每对有奸情的人a与b,因为它们不能同时在新娘对面(右边),所以:

a*2+1->b*2

b*2+1->a*2

        注意首先我们定了新娘(0号)在左边,新郎(第1号人)一定在右边,所以我们要先加上:

0*2+1->0*2  和 1*2->1*2+1 

这样就保证了新娘和新郎在固定的那边不动.

AC代码:

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 1000*2+100;
struct TwoSAT
{
    int n;
    vector<int> G[maxn*2];
    int S[maxn*2],c;
    bool mark[maxn*2];

    bool dfs(int x)
    {
        if(mark[x^1]) return false;
        if(mark[x]) return true;
        mark[x]=true;
        S[c++]=x;
        for(int i=0;i<g if return false true void init n this->n=n;
        for(int i=0;i0) mark[S[--c]]=false;
                if(!dfs(i+1)) return false;  //注意细节,这里写成了return true;
            }
        }
        return true;
    }
}TS;
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        if(n==0&&m==0) break;
        TS.init(n*2);
        TS.add_clause(0,1,0,0);//新娘放左
        TS.add_clause(1,0,1,1);//新郎放右
        for(int i=1;i<n int a="i*2;" b="i*2+1;//丈夫" ts.add_clause for i="0;i<m;i++)" char s1 scanf if else printf luck return><br>
<br>


</n></g></int></algorithm></vector></cstring></cstdio>
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