poj 3311 Hie with the Pie (TSP问题)

Hie with the Pie

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4491		Accepted: 2376

Description

The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can afford to hire only one driver to do the deliveries. He will wait for 1 or more (up to 10) orders to be processed before
he starts any deliveries. Needless to say, he would like to take the shortest route in delivering these goodies and returning to the pizzeria, even if it means passing the same location(s) or the pizzeria more than once on the way. He has commissioned you
to write a program to help him.

Input

Input will consist of multiple test cases. The first line will contain a single integer n indicating the number of orders to deliver, where 1 ≤ n ≤ 10. After this will be n + 1 lines each containing n + 1 integers indicating
the times to travel between the pizzeria (numbered 0) and the n locations (numbers 1 to n). The jth value on the ith line indicates the time to go directly from location i to location j without visiting
any other locations along the way. Note that there may be quicker ways to go from i to j via other locations, due to different speed limits, traffic lights, etc. Also, the time values may not be symmetric, i.e., the time to go directly from
location i to j may not be the same as the time to go directly from location j to i. An input value of n = 0 will terminate input.

Output

For each test case, you should output a single number indicating the minimum time to deliver all of the pizzas and return to the pizzeria.

Sample Input

3
0 1 10 10
1 0 1 2
10 1 0 10
10 2 10 0
0

Sample Output

8

题意：从起点0開始。遍历全部的点并回到起点的最短距离。每一个点能够经过多次。

ans1:

DP+状态压缩：dp[state][i]表示起点到达i点,状态为state的最短距离。

dp[state][i] =min{dp[state][i],dp[state'][j]+dis[j][i]} dis[j][i]为j到i的最短距离；

</pre><pre name="code" class="cpp">#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<queue>
using namespace std;
#define ll __int64
#define mem(a,t) memset(a,t,sizeof(a))
#define N 12

const int inf=0x3fffffff;
int g[N][N];
int dp[1<<11][N];//dp[state][i]表示在当前状态下，从起点0到达i点的最短距离
void floyd(int n)
{
    int i,j,k;
    for(k=0; k<=n; k++)   //Floyd求最短路
        for(i=0; i<=n; i++)
            for(j=0; j<=n; j++)
                g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
}
int main()
{
    //freopen("in.txt","r",stdin);
    int n,i,j,k,s;
    while(scanf("%d",&n),n)
    {
        for(i=0; i<=n; i++)
            for(j=0; j<=n; j++)
                scanf("%d",&g[i][j]);
        floyd(n);
        for(i=0;i<(1<<(n+1));i++)
            for(j=0;j<=n;j++)
            dp[i][j]=inf;
        dp[0][0]=0;
        for(s=0; s<(1<<(n+1)); s++) //枚举每一个状态
        {
            for(i=0; i<=n; i++)     //枚举中间点。看能否使距离变短
            {
                if(!(s&(1<<i)))
                    continue;
                for(k=0; k<=n; k++)
                    dp[s][i]=min(dp[s][i],dp[s^(1<<i)][k]+g[k][i]);
            }
        }
        printf("%d\n",dp[(1<<(n+1))-1][0]);
    }
    return 0;
}

ans2:

bfs+状态压缩：先用Floyd求出随意两点之间的最短路。然后。能够用广搜求得答案，搜索中的每一个点第一次到达用二进制位进行标记。

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
#define N 12
const int inf=0x3fffffff;
struct node
{
    int x,s,t;  //位置、状态、时间
    int cnt;    //訪问地点数目
    friend bool operator<(node a,node b)
    {
        return a.t>b.t;
    }
};
int mark[N][1030];
int g[N][N];
int n,ans;
void Floyd()
{
    int i,j,k;
    for(k=0;k<=n;k++)
    {
        for(i=0;i<=n;i++)
        {
            for(j=0;j<=n;j++)
            {
                g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
            }
        }
    }
}
void bfs(int u)
{
    int i;
    priority_queue<node >q;
    node cur,next;
    cur.x=u;
    cur.t=cur.s=cur.cnt=0;
    q.push(cur);
    mark[u][0]=0;
    q.push(cur);
    while(!q.empty())
    {
        cur=q.top();
        q.pop();
        for(i=0;i<=n;i++)
        {
            next.s=cur.s;
            next.t=cur.t;
            next.cnt=cur.cnt;
            next.x=i;
            next.t+=g[cur.x][i];
            if(i&&(next.s&(1<<(i-1)))==0)
            {
                next.s|=(1<<(i-1));
                next.cnt++;
            }
            if(next.t<mark[i][next.s])
            {
                mark[i][next.s]=next.t;
                if(next.cnt==n)
                {
                    ans=min(ans,next.t+g[i][0]);
                    continue;
                }
                q.push(next);
            }
        }
    }
}
int main()
{
    int i,j;
    while(scanf("%d",&n),n)
    {
        for(i=0;i<=n;i++)
        {
            for(j=0;j<=n;j++)
            {
                scanf("%d",&g[i][j]);
            }
        }
        Floyd();
        for(i=0;i<=n;i++)
        {
            for(j=0;j<(1<<n);j++)
                mark[i][j]=inf;
        }
        ans=inf;
        bfs(0);
        printf("%d\n",ans);
    }
    return 0;
}