HDU3538 A sample Hamilton path

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 503    Accepted Submission(s): 200

Problem Description

Give you a Graph,you have to start at the city with ID zero.

 

Input

The first line is n(1<=n<=21) m(0<=m<=3)
The next n line show you the graph, each line has n integers.
The jth integers means the length to city j.if the number is -1 means there is no way. If i==j the number must be -1.You can assume that the length will not larger than 10000
Next m lines,each line has two integers a,b (0<=a,b<n) means the path must visit city a first.
The input end with EOF.

 

Output

For each test case,output the shorest length of the hamilton path.
If you could not find a path, output -1

 

Sample Input

3 0 -1 2 4 -1 -1 2 1 3 -1 4 3 -1 2 -1 1 2 -1 2 1 4 3 -1 1 3 2 3 -1 1 3 0 1 2 3

 

Sample Output

4 5

Hint

I think that all of you know that a!=b and b!=0 =。=

 

Source

2010 ACM-ICPC Multi-University Training Contest（11）——Host by BUPT

 

Recommend

zhouzeyong

 

 

状态压缩DP，详解见代码

/**/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=1e6;
const int mxn=4194304;//2^22
int dp[mxn][22];//[遍历状态][最后到达点]=最短路径 
int dis[22][22];
int pre[22];//每个点的前驱要求 
int n,m;
int xn;
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        memset(pre,0,sizeof pre);
        int i,j;
        xn=1<<n;
        for(i=1;i<xn;i++)
         for(j=0;j<n;j++){
             dp[i][j]=INF;
         }
        //init
        for(i=0;i<n;i++)
         for(j=0;j<n;j++){
             scanf("%d",&dis[i][j]);
             if(dis[i][j]==-1)dis[i][j]=INF;
         }
        int u,v;
        for(i=1;i<=m;i++){//保存前驱要求 
            scanf("%d%d",&u,&v);
            pre[v]|=(1<<u);
        }
        dp[1][0]=0;
        for(i=1;i<xn;i++){
            for(j=0;j<n;j++){
                if(dp[i][j]==INF)continue;//i状态之前没走到 
                for(int k=1;k<n;k++){
                    if(!(i&(1<<j)))continue;//j不在已走过的集合中
                    if(i&(1<<k))continue;//k在走过的集合中
                    if(pre[k]!=(i&pre[k]))continue;//k点前驱要求未满足 
                    dp[i|(1<<k)][k]=min(dp[i|(1<<k)][k],dp[i][j]+dis[j][k]);
                }
            }
        }
        int ans=INF;
        for(i=0;i<n;i++)ans=min(ans,dp[xn-1][i]);
        if(ans>=INF) printf("-1
");
        else printf("%d
",ans);
    }
    return 0;
}