状压DP HDU3538 A sample Hamilton path

A sample Hamilton path

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 527    Accepted Submission(s): 213

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

 

莫名其妙就状压DP了，说实在我现在还不明白这是个什么玩意儿......

不过尹神说了是典型，那就写一写吧......

这就是个Floyd么......

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int n,m;
int dis[25][25];
int f[2500000];
int dp[2500000][25];//dp[i][j]已访问的点集i 到达点i的最短hamilton 
const int MAX=3000000;
int ans;

int main(){
    int x=0,y=0;
    while(scanf("%d%d",&n,&m)!=EOF){
        memset(f,0,sizeof(f));
        memset(dis,0,sizeof(dis));
        memset(dp,0,sizeof(dp)); 
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++) scanf("%d",&dis[i][j]);
        for(int i=1;i<=m;i++){
            scanf("%d%d",&x,&y);
            f[y]|=(1<<x);//观察了样例才发现 原来y可以多次出现 
        }
        for(int i=0;i<(1<<n);i++)
            for(int j=0;j<n;j++)  dp[i][j]=MAX;
        dp[1][0]=0;
        for(int k=0;k<(1<<n);k++)
            for(int i=0;i<n;i++)
                if(dp[k][i]!=MAX)
                    for(int j=0;j<n;j++){
                        if((dis[i][j]==-1)||(!(k&(1<<i)))||(k&(1<<j))||f[j]!=(k&f[j])) continue;//判断是否遍历过 
                        dp[k|(1<<j)][j]=min(dp[k|(1<<j)][j],dp[k][i]+dis[i][j]);//f[k][i]->f[k+{j}][j] + dis(i, j) 
                    }
        ans=MAX;
        for(int i=0;i<n;i++) ans=min(ans,dp[(1<<n)-1][i]);
        if(ans==MAX) printf("-1
");
        else printf("%d
",ans);
    }
    return 0;
}