HDU 5418 Victor and World State compression DP, travel business issue

Victor and World Time limit:4000/2000 MS (java/others) Memory limit:262144/131072 K (java/others)
Total submission (s): 385 Accepted Submission (s): 160


Problem Description After trying-many years, Victor has finally received a pilot license. To has a celebration, he intends to buy himself an airplane and fly around the world. There is n countries on the Earth, which is numbered from 1 to N. They is connected by M undirected flights, detailedly the i-th flight connects the ui-th and the vi-th country, and I T would 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 was at the country whose number was 1, he wants to know the minimal amount of fuel for him to visit every countr Y 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 the test cases.
In every test case, there is, integers n and m in the first line, denoting the number of the countries and the Numbe R of the flights.

Then there is 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 Amoun T of fuel for Victor to finish the travel.
Sample Input

1 3 2 1 2 2 1 3 3
Sample Output

10
Source bestcoder Round #52 (Div.2)

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace
Std
int map[20][20];

int dp[1<<17][20];
    int main () {int t,n,m,w,u,v;
    scanf ("%d", &t);
        while (t--) {scanf ("%d%d", &n,&m);
        memset (map,0x3f,sizeof (map));
        for (int i = 0;i < n; i++) map[i][i] = 0;
            for (int i = 0;i < m; i++) {scanf ("%d%d%d", &u,&v,&w);
            U--, v--;
        MAP[U][V] = map[v][u] = min (map[u][v],w); } for (int i = 0;i < n;i++) for (int j = 0;j < N; j + +) for (int k = 0;k < n; k +
        +) Map[j][k] = map[k][j] = min (map[j][k],map[j][i]+map[i][k]);
        Memset (Dp,0x3f,sizeof (DP));
        Dp[1][0] = 0;
                    for (int i = 1;i < (1<<n), i++) {for (int j = 0;j < N; j + +) {if (i& (1<<j)) {
            for (int k = 0;k < n; k++) {            if ((i& (1<<k)) ==0 && J! = k) {dp[i| ( 1<<k)][k] = min (dp[i| (
                        1<<k)][k],dp[i][j]+map[j][k]);
        }}}}} int ans = dp[0][0];
        for (int i = 0;i < n; i++) {ans = min (ans,dp[(1<<n) -1][i]+map[i][0]);
    } printf ("%d\n", ans);
} return 0;




















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