Hie with the Pie
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 4491 | Accepted: 2376 |
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
思路:先用Floyd求出任意两点之间的最短路，然后，可以用广搜求得答案，搜索中的每个点第一次到达用二进制位进行标记。
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
#define N 12
const int inf=0x3fffffff;
struct node
{
int x,s,t; //位置、状态、时间
int cnt; //访问地点数目
friend bool operator<(node a,node b)
{
return a.t>b.t;
}
};
int mark[N][1030];
int g[N][N];
int n,ans;
void Floyd()
{
int i,j,k;
for(k=0;k<=n;k++)
{
for(i=0;i<=n;i++)
{
for(j=0;j<=n;j++)
{
g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
}
}
}
}
void bfs(int u)
{
int i;
priority_queue<node >q;
node cur,next;
cur.x=u;
cur.t=cur.s=cur.cnt=0;
q.push(cur);
mark[u][0]=0;
q.push(cur);
while(!q.empty())
{
cur=q.top();
q.pop();
for(i=0;i<=n;i++)
{
next.s=cur.s;
next.t=cur.t;
next.cnt=cur.cnt;
next.x=i;
next.t+=g[cur.x][i];
if(i&&(next.s&(1<<(i-1)))==0)
{
next.s|=(1<<(i-1));
next.cnt++;
}
if(next.t<mark[i][next.s])
{
mark[i][next.s]=next.t;
if(next.cnt==n)
{
ans=min(ans,next.t+g[i][0]);
continue;
}
q.push(next);
}
}
}
}
int main()
{
int i,j;
while(scanf("%d",&n),n)
{
for(i=0;i<=n;i++)
{
for(j=0;j<=n;j++)
{
scanf("%d",&g[i][j]);
}
}
Floyd();
for(i=0;i<=n;i++)
{
for(j=0;j<(1<<n);j++)
mark[i][j]=inf;
}
ans=inf;
bfs(0);
printf("%d\n",ans);
}
return 0;
}