poj 3311 Hie with the Pie

最新推荐文章于 2025-09-04 15:42:34 发布

疯狂的橡树

最新推荐文章于 2025-09-04 15:42:34 发布

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分类专栏：动态规划文章标签：算法 C ACM 动态规划

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本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了视频分割的基本概念、算法及其应用，并详细阐述了语义识别在智能视频分析中的重要作用，包括其在自动驾驶、AR增强现实等场景中的实际应用。

Hie with the Pie

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3017		Accepted: 1540

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

Sample Output

Source

East Central North America 2006

状态压缩。找到状态转移方程后很简单。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int INF=0x3f3f3f3f;
int dis[15][15];//记录两点之间的最短距离
int dp[1<<11][12];//dp[i][j]由状态i到j点的最短距离。i中的0和1代表是否走过
int n;
int mi(int a,int b){ return a<b?a:b; }
void floyd()//计算两点间的最短距离
{
    int i,j,k;

    for(i=1;i<=n;i++)
        for(k=1;k<=n;k++)
           for(j=1;j<=n;j++)
           if(dis[i][k]+dis[k][j]<dis[i][j])
           dis[i][j]=dis[i][k]+dis[k][j];
}
int main()
{
    int i,j,k,next,ans;

    while(scanf("%d",&n),n)
    {
        n++;
        memset(dp,0x3f,sizeof dp);
        memset(dis,0x3f,sizeof dis);
        dp[1][1]=0;
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            scanf("%d",&dis[i][j]);
        floyd();
        for(i=0;i<(1<<n);i++)//枚举到j点时的状态
        {
            for(j=1;j<=n;j++)
            {
                if(dp[i][j]==INF)//到j点后状态刚好变为i
                    continue;
                for(k=1;k<=n;k++)//ｊ为中间点
                {
                    if(j==k)//如果已经到过k
                        continue;
                    if(i&(1<<(k-1)))//如果来过说明是第二次来。没有第一次来距离短
                        continue;
                    next=i|(1<<(k-1));//记录到k后的状态
                    dp[next][k]=mi(dp[next][k],dp[i][j]+dis[j][k]);
                }
            }
        }
        ans=INF;
        for(i=1;i<=n;i++)//找到到达目的状态(全为1)后的最短距离
            ans=mi(ans,dp[(1<<n)-1][i]+dis[i][1]);
        printf("%d\n",ans);
    }
    return 0;
}