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 integernindicating the number of orders to deliver, where 1 ≤n≤ 10. After this will ben+ 1 lines each containingn+ 1 integers indicating the times to travel between the pizzeria (numbered 0) and thenlocations (numbers 1 ton). Thejth value on theith line indicates the time to go directly from locationito locationjwithout visiting any other locations along the way. Note that there may be quicker ways to go fromitojvia 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 locationitojmay not be the same as the time to go directly from locationjtoi. An input value ofn= 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
30 1 10 101 0 1 210 1 0 1010 2 10 00
Sample Output
8
由于路径的不确定性，要floyd先求一下最短路然后就是状压。
最后由于要返回店里，，所以要加上到原点（0）的距离（最短）。
#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<set>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;const int INF=0x3f3f3f3f;typedef pair<int,int>pil;int mp[20][20];int dp[1<<12][12];int n;void floyed(){for(int k=0;k<=n;k++){for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);}}int main(){while(~scanf("%d",&n)&&n){REPF(i,0,n)REPF(j,0,n) scanf("%d",&mp[i][j]);CLEAR(dp,INF);floyed();n++;dp[0][0]=0;int status=(1<<n)-1;for(int i=0;i<=status;i++){for(int j=0;j<n;j++){if(dp[i][j]==INF) continue;for(int k=0;k<n;k++){if(i&(1<<k)) continue;int st=i^(1<<k);dp[st][k]=min(dp[st][k],dp[i][j]+mp[j][k]);}}}int ans=INF;for(int i=0;i<n;i++)ans=min(ans,dp[status][i]+mp[i][0]);printf("%d\n",ans);}return 0;}
不付出，却一定不会有收获，不要奢望出现奇迹。
