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
Solution
好氣哦，調了好長時間還重寫了一遍
結果發現送完還要回到披薩店…我的閱讀理解能力可能出現了障礙
從披薩店出發，要經過每個送餐點（可以重複經過），再回到披薩店，求總共需要的最少時間
Floyd處理出所有最短路
f數組第一維記錄每個地點是否被經過的狀態，第二維記錄此時所在的地點
init中的處理使得state數組記錄的狀態相當於是按照去到的地方從少到多排序的
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cstdlib>
#define Min(a,b) (a<b?a:b)
#define inf 0x3f3f3f3f
using namespace std;
int n,dis[11][11],state[1100],tot,s[11][1100],num[11];
int f[1100][11];
int count(int x)
{
int cnt=0;
while(x)
{
if(x&1)cnt++;
x>>=1;
}
return cnt;
}
void init()
{
memset(num,0,sizeof(num));
memset(f,0x3f,sizeof(f));
tot=0;
for(int i=0;i<(1<<n);i++)
{
int x=count(i);
s[x][++num[x]]=i;
if(x==1)f[i][num[1]]=dis[0][num[1]];
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=num[i];j++)
{
state[++tot]=s[i][j];
}
}
}
int main()
{
while(~scanf("%d",&n)&&n)
{
memset(dis,0x3f,sizeof(dis));
for(int i=0;i<=n;i++)
for(int j=0;j<=n;j++)
scanf("%d",&dis[i][j]);
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]);
init();
for(int i=1;i<=tot;i++)
{
for(int j=1;j<=n;j++)
{
if(!(state[i]&(1<<(j-1))))continue;
int t=state[i]^(1<<(j-1));
for(int k=1;k<=n;k++)
{
if(!(t&(1<<(k-1))))continue;
f[state[i]][j]=Min(f[state[i]][j],f[t][k]+dis[k][j]);
}
}
}
int ans=inf;
for(int i=1;i<=n;i++)
ans=Min(ans,f[(1<<n)-1][i]+dis[i][0]);
printf("%d\n",ans);
}
return 0;
}