Hie with the Pie

題目鏈接

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

題目大意

題目大概的意思就是有n個位置,需要送上pie,之后回到起點,這就是旅行商變形后的問題,其實一個點會經過很多次,但是核心還是遍歷所有的點.我們用floyd求出相互之間的最短路徑.樣例的路徑為0->1->2->3->0,他們之間的最短路為
1 1 3 3 為 8.
我們狀態壓縮來表示一個點是否走過.dp[s][i]表示,已經走過了s中的點,最后一個經過的點為 i .s為二進制表示的一個狀態,當s全為1的時候表示全部走完.遍歷狀態的時候,我們枚舉每一種狀態,dp[s][i]可以從未含有i的dp[s`][j]中轉移,轉移過程需要加上j到i的距離.
參考題解

代碼實現

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=25; const int inf=0x3f3f3f3f; int map[maxn][maxn]; int n; int dp[1<<22][maxn]; int main() {while(cin>>n&&n){for(int i=0;i<=n;i++){for(int j=0;j<=n;j++)cin>>map[i][j];}for(int k=0;k<=n;k++)for(int i=0;i<=n;i++)for(int j=0;j<=n;j++) map[i][j]=min(map[i][j],map[i][k]+map[k][j]);int m=(1<<n)-1;for(int s=0;s<=m;s++){for(int i=1;i<=n;i++){if(s&(1<<i-1)){if(s==(1<<(i-1)))dp[s][i]=map[0][i];else{dp[s][i]=inf;for(int j=1;j<=n;j++){if(s&(1<<(j-1))&&j!=i){dp[s][i]=min(dp[s][i],dp[s-(1<<(i-1))][j]+map[j][i]);//從中拿掉i點,從未擁有i的集合中更新}}}}}}int ans=inf;for(int i=1;i<=n;i++)ans=min(ans,dp[m][i]+map[i][0]);//判斷回到起點后的總距離cout<<ans<<endl;} }

總結

以上是生活随笔為你收集整理的Hie with the Pie(旅行商问题)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。