题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=5418
Victor and World
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 958 Accepted Submission(s): 431
Problem Description
After trying hard for many years, Victor has finally received a pilot license. To have a celebration, he intends to buy himself an airplane and fly around the world. There are n countries on the earth, which are numbered from 1 to n. They are connected by m undirected flights, detailedly the i-th flight connects the ui-th and the vi-th country, and it will cost Victor‘s airplane wi L fuel if Victor flies through it. And it is possible for him to fly to every country from the first country.
Victor now is at the country whose number is 1, he wants to know the minimal amount of fuel for him to visit every country at least once and finally return to the first country.
Input
The first line of the input contains an integer T, denoting the number of test cases.
In every test case, there are two integers n and m in the first line, denoting the number of the countries and the number of the flights.
Then there are m lines, each line contains three integers ui, vi and wi, describing a flight.
1≤T≤20.
1≤n≤16.
1≤m≤100000.
1≤wi≤100.
1≤ui,vi≤n.
Output
Your program should print T lines : the i-th of these should contain a single integer, denoting the minimal amount of fuel for Victor to finish the travel.
Sample Input
1
3 2
1 2 2
1 3 3
Sample Output
10
Source
题意： 给一个地图，从其中一个点开始走，遍历完所有的点后最后再回到这个点，求最短路径
题解： 注意数据范围给的是16个点，又是求汉密顿环路问题，就想到了状态压缩dp，这个题可以学习一个很好地思想就是spfa的思想，将其用到dp中，取出队首的状态，看这个状态停在哪个点，用这个点更新所有的可以更新的状态，因为每一个点可以多次的遍历，所以每次更新都要从0到n依次遍历，新更新的状态如果之前没有访问过，则说明这个状态有更新其他状态的潜力，所以将其压入队列中，通过这种方式可以更新所有的状态。
简单说一下状态压缩dp，用一个二进制的数表示某个点是否在集合内，全集为(1<<n)-1 ; j集合中加入一个元素k是 j|(1<<k) ; 在j集合中去掉一个元素k是 j^(1<<k) ;
dp[j][k] 表示走过集合j中的所有元素最后停留在k点的最短路径
转移方程： dp[s|(1<<i)][i] = min(dp[s|(1<<i)][i] , dp[s][u]+mp[u][i])
注意事项： 这里的边是双向边，在输入边的时候会有很多的无效边，要取最小的。
用qair<int , int> 相当于一个struct{ int a, int b};
代码：
1 #include<cstdio>
2 #include<cstring>
3 #include<queue>
4 #include<algorithm>
5 using namespace std;
6
7 int dp[1<<17][17];
8 int mp[17][17];
9 int dis[17];
10 bool vis[1<<17][17];
11 int n;
12
13 int main()
14 {
15 int T , m ;
16 scanf("%d",&T);
17 while(T--)
18 {
19 scanf("%d%d",&n,&m);
20 memset(mp,0x7f,sizeof(mp));
21 for(int i =0 ; i < n ;i++) mp[i][i] = 0;
22 for(int i =0 ; i < m ;i++) {
23 int u , v , d;
24 scanf("%d%d%d",&u,&v,&d);
25 u--,v--;
26 mp[u][v] = mp[v][u] = min(mp[u][v],d);
27 }
28 memset(dp,0x7f,sizeof(dp));
29 memset(vis,0,sizeof(vis));
30 dp[0][0] = 0 ; vis[0][0] = 1;
31 queue<pair<int,int> > q;
32 q.push(make_pair(0,0));
33 while(!q.empty())
34 {
35 int s = q.front().first;
36 int u = q.front().second;
37 q.pop();
38 for(int i = 0 ; i < n ; i++)
39 {
40 int ss = s|(1<<i);
41 if(dp[ss][i]>dp[s][u] + mp[u][i]){
42 dp[ss][i] = dp[s][u] + mp[u][i];
43 if(vis[ss][i] == 0){
44 vis[ss][i] = 1 ;
45 q.push(make_pair(ss,i));
46 }
47 }
48 }
49 }
50 printf("%d\n",dp[(1<<n)-1][0]);
51 }
52 return 0 ;
53 }