CF11D A Simple Task(状压DP)

\(solution:\)

思路大家应该都懂:

状压DP:\(f[i][j]\),其中 \(i\) 这一维是需要状压的,用来记录19个节点每一个是否已经走过(走过为 \(1\) ,没走为 \(0\) ,用 \(2\)进制 压缩一下即可)。同时,我们认为状压中已经走过的序号最小的节点为出发节点,\(j\) 即数组第二维是路径终点。(当这两个数相同时,说明找到了一个环)。

注:这种方法因为无向图的存在,会出现(同一条路径出现两次)(一条边和两个端点构成非法环)的情况,这只需要在输出答案时 \(ans=(ans-m)/2\) 即可!

\(code:\)

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>

#define ll long long
#define db double
#define inf 0x7fffffff
#define rg register int

using namespace std;

int n,m,t,u,v;
bool a[19][19]; //存边
ll ans,f[600001][19]; //状压

inline int qr(){ //快读
    char ch;
    while((ch=getchar())<'0'||ch>'9');
    int res=ch^48;
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+(ch^48);
    return res;
}

int main(){
    //freopen("hamilton.in","r",stdin);
    //freopen("hamilton.out","w",stdout);
    n=qr(),m=qr();t=1<<n;
    for(rg i=1;i<=m;++i){
        u=qr()-1;v=qr()-1;
        a[u][v]=a[v][u]=1;//加边
    }
    for(rg i=0;i<n;++i)
        f[1<<i][i]=1; //初始化(创建以i为起点的路径)
    for(rg i=1;i<=t;++i){
        for(rg j=0;j<n;++j){
            if(!f[i][j])continue; //加速
            for(rg k=0;k<n;++k){
                if(!a[j][k])continue; //加速
                if((i&-i)>1<<k)continue; //起点不能改!!!(去重)
                if(1<<k&i){ //这个点走过
                    if(1<<k==(i&-i)) //起点与终点相同
                        ans+=f[i][j];
                }else f[i|1<<k][k]+=f[i][j]; //没走过就走!
            }
        }
    }printf("%lld",(ans-m)/2); //处理之后再输出!
    return 0;
}

原文地址:https://www.cnblogs.com/812-xiao-wen/p/10322546.html

时间: 2024-10-15 20:25:50

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