題目：
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
代碼如下：

#include<iostream> using namespace std; int n,a[15][15],dp[3000][15]; int dis[15][15]; void floyd() {for(int k = 0;k <= n;k++)for(int i = 0;i <= n;i++)for(int j = 0;j <= n;j++)dis[i][j] = min(dis[i][j],dis[i][k] + dis[k][j]); } int main() {while(~scanf("%d",&n) && n != 0){for(int i = 0;i <= n;i++) for(int j = 0;j <= n;j++) cin >> dis[i][j];floyd();for(int s = 0;s < (1 << n);s++){for(int i = 1;i <= n;i++){ //從第1個(gè)點(diǎn)到第n個(gè)點(diǎn)嘗試 if(s & (1 << (i - 1))){ //狀態(tài)s中已經(jīng)走到i點(diǎn) if(s == (1 << (i - 1))) dp[s][i] = dis[0][i];//從0點(diǎn)出發(fā)走到i點(diǎn)，dp的邊界else{dp[s][i] = 0x3f3f3f3f;for(int j = 1;j <= n;j++){if(i != j && s & (1 << (j - 1))) //j不能等于i,并且狀態(tài)s中要走到j(luò)點(diǎn) dp[s][i] = min(dp[s][i],dp[s ^ (1 << (i - 1))][j] + dis[j][i]);//松弛操作 }} } }} int ans = 0x3f3f3f3f;for(int i = 1;i <= n;i++) ans = min(ans,dp[((1 << n) - 1)][i] + dis[i][0]);cout << ans << endl; }return 0; }

題意：
從0點(diǎn)出發(fā)經(jīng)過(guò)每個(gè)點(diǎn)再回到0點(diǎn)的最短距離。
思路：
如果暴力枚舉會(huì)發(fā)現(xiàn)會(huì)有很多重疊的子問(wèn)題，所以這里采用狀壓dp來(lái)做。因?yàn)橐傈c(diǎn)和點(diǎn)之間的最短距離，所以這里先要用floyd算法計(jì)算出點(diǎn)與點(diǎn)之間的最短路徑。然后進(jìn)行dp，dp[s][i]代表狀態(tài)為s時(shí)正好走到i點(diǎn)所走的路徑。具體過(guò)程代碼旁有注釋。

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