Shopping
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 632 Accepted Submission(s): 209
Problem Description
You have just moved into a new apartment and have a long list of items you need to buy. Unfortunately, to buy this many items requires going to many different stores. You would like to minimize the amount of driving necessary to buy all the items you need.
Your city is organized as a set of intersections connected by roads. Your house and every store is located at some intersection. Your task is to find the shortest route that begins at your house, visits all the stores that you need to shop at, and returns to your house.
Input
The first line of input contains a single integer, the number of test cases to follow. Each test case begins with a line containing two integers N and M, the number of intersections and roads in the city, respectively. Each of these integers is between 1 and 100000, inclusive. The intersections are numbered from 0 to N-1. Your house is at the intersection numbered 0. M lines follow, each containing three integers X, Y, and D, indicating that the intersections X and Y are connected by a bidirectional road of length D. The following line contains a single integer S, the number of stores you need to visit, which is between 1 and ten, inclusive. The subsequent S lines each contain one integer indicating the intersection at which each store is located. It is possible to reach all of the stores from your house.
Output
For each test case, output a line containing a single integer, the length of the shortest possible shopping trip from your house, visiting all the stores, and returning to your house.
Sample Input
1
4 6
0 1 1
1 2 1
2 3 1
3 0 1
0 2 5
1 3 5
3
1
2
3
Sample Output
4
Source
University of Waterloo Local Contest 2010.07.10
思路:给你一个无向图，求从0号点开始遍历所有的指定点再回到0号点的最短路径，指定点只有10个，所以先spfa后dfs即可;
#include <iostream>
#include <set>
#include <queue>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int MAX=101000;
const int INF = 0x3f3f3f3f;
typedef struct node
{
int v;
int w;
int next;
}Edge;
int Head[MAX];
Edge Line[MAX*2];
int top;
int n,m,s,Map[15][15];
int dis[12][MAX],num[15];
bool vis[MAX];
bool vist[15];
void AddEdge(int u,int v,int w)//由于点的个数比较多,所以采用领接表的形式
{
Line[top].v=v;
Line[top].w=w;
Line[top].next=Head[u];
Head[u]=top++;
}
void spfa(int s,int site)//spfa计算出要去的商店之间的距离
{
memset(vis,false,sizeof(false));
for(int i=0;i<=n;i++)
{
dis[site][i]=INF;
}
dis[site][s]=0;
vis[s]=true;
queue<int>Q;
Q.push(s);
while(!Q.empty())
{
int u=Q.front();
Q.pop();
for(int i=Head[u];i!=-1;i=Line[i].next)
{
if(dis[site][Line[i].v]>dis[site][u]+Line[i].w)
{
dis[site][Line[i].v]=dis[site][u]+Line[i].w;
if(!vis[Line[i].v])
{
Q.push(Line[i].v);
vis[Line[i].v]=true;
}
}
}
vis[u]=false;
}
}
int dfs(int st,int sum,int num)//搜索出最小的距离
{
vist[st]=true;
if(st==0)
{
if(num==s+1)
{
return sum;
}
else
{
return INF;
}
}
int ans=INF;
for(int i=0;i<=s;i++)
{
if(!vist[i]||i==0)
{
ans=min(ans,dfs(i,sum+Map[st][i],num+1));
}
}
vist[st]=false;
return ans;
}
int main()
{
int T;
int u,v,w;
scanf("%d",&T);
while(T--)
{
memset(Head,-1,sizeof(Head));
scanf("%d %d",&n,&m);
top=0;
for(int i=0;i<m;i++)
{
scanf("%d %d %d",&u,&v,&w);
AddEdge(u,v,w);
AddEdge(v,u,w);
}
num[0]=0;
spfa(0,0);
scanf("%d",&s);
for(int i=1;i<=s;i++)
{
scanf("%d",&num[i]);
spfa(num[i],i);
}
memset(Map,0,sizeof(Map));
for(int i=0;i<=s;i++)//离散化,由于要去的点比较少,可以将要去的点重新进行建图
{
for(int j=0;j<=s;j++)
{
Map[i][j]=dis[i][num[j]];
}
}
int ans=INF;
for(int i=1;i<=s;i++)
{
memset(vist,false,sizeof(vist));
vist[0]=true;
ans=min(ans,dfs(i,Map[0][i],1));
}
printf("%d
",ans);
}
return 0;
}