描述

傳送門：poj-3311

?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.

Examples

• intput

  1 2 3 4 5 6

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

• output

  1

  8

題目大意

給你一個(gè)有$n+1$，$(1<=n<=10)$個(gè)點(diǎn)的有向完全圖，用矩陣的形式給出任意兩個(gè)不同點(diǎn)之間的距離。（其中從$i$到$j$的距離不一定等于從$j$到$i$的距離）現(xiàn)在要你求出從$0$號(hào)點(diǎn)出發(fā)，走過$1$到$n$號(hào)點(diǎn)至少一次，然后再回到$0$號(hào)點(diǎn)所花的最小時(shí)間。

思路

• 先用floyd $O(n^3)$預(yù)處理出任意兩個(gè)點(diǎn)之間的最短距離。
• 用一個(gè)11位的二進(jìn)制數(shù)表示狀態(tài)，1表示要經(jīng)過這個(gè)點(diǎn)，0表示不經(jīng)過。
• $dp[state][j]$: 在$state$狀態(tài)下$0$點(diǎn)到達(dá)$j$點(diǎn)的最短路。如果state狀態(tài)里面經(jīng)過了i，并且沒有經(jīng)過j，那么就能通過dp[state][i]更新dp[state|(1<<j)][j]。
• 這道題的狀態(tài)和狀態(tài)轉(zhuǎn)移方程和zoj-3471很像：

代碼

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52

/* Problem: 3311 User: Armin Memory: 396K Time: 0MS Language: G++ Result: Accepted */ #include<stdio.h> #include<algorithm> using namespace std; #define rep(i,a,n) for(int i=a;i<n;i++) #define repd(i,a,n) for(int i=n-1;i>=a;i--) #define CRL(a,x) memset(a,x,sizeof(a)) #define Max 0x3f3f3f const int N=11; int Map[N][N],dp[1<<N][N],n; int main() { while(scanf("%d",&n),n++){ rep(i,0,n) rep(j,0,n) scanf("%d",&Map[i][j]); rep(k,0,n) //floy預(yù)處理出任意兩個(gè)點(diǎn)之間的最短距離 rep(i,0,n) rep(j,0,n) if(Map[i][j]>Map[i][k]+Map[k][j]) Map[i][j]=Map[i][k]+Map[k][j]; rep(i,0,1<<n) //初始化成無窮大 rep(j,0,n) dp[i][j]=Max; dp[0][0]=0; //0點(diǎn)在0狀態(tài)到0點(diǎn)的距離為0 rep(state,0,1<<n) rep(i,0,n){ if(dp[state][i]==Max) continue; //剪枝：只有當(dāng)dp[state][i]不等于Max時(shí)才可能更新 if(i&&state==(1<<i)) //如果state狀態(tài)只經(jīng)過了i一個(gè)點(diǎn)，注意這里要把i==0的情況去點(diǎn)，因?yàn)閕==0在33行已經(jīng)處理了，這里1<<0 ==1 !=0 所以這里里更新會(huì)出錯(cuò)。 dp[state][i]=Map[0][i]; else rep(j,0,n) if(((1<<j)&state)==0) //如果沒經(jīng)過j城市 dp[state|(1<<j)][j]=min(dp[state|(1<<j)][j],dp[state][i]+Map[i][j]); } int ans=Max; rep(i,1,n) ans=min(ans,dp[(1<<n)-1][i]+Map[i][0]); //找最優(yōu)解+回來的路 printf("%d\n",ans); } return 0; }

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