Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem A: Shopping

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 Specification

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.

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

Output Specification

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.

Output for Sample Input

4

---

#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#define none -1
using namespace std;
typedef struct {
    int to, v;
} Arc;
typedef vector<Arc>::iterator it;
vector<Arc>  link[100000];
void SPFA(int n, int st, int dis[]) {
    queue<int> Q;
    bool used[n];
    memset(used, 0, sizeof(used));
   
    for(it i = link[st].begin(); i != link[st].end(); i++) {
        if(dis[i->to] > i->v) {
            dis[i->to] = i->v;
            if(!used[i->to]) {
                used[i->to] = true;
                Q.push(i->to);
            }
        }
    }
    int tv;
    while(!Q.empty()) {
        tv = Q.front();
        Q.pop();
        used[tv] = false;
        for(it i = link[tv].begin(); i != link[tv].end(); i++) {
            if(dis[i->to] > dis[tv] + i->v) {
                dis[i->to] = dis[tv] + i->v;
                if(!used[i->to]) {
                    used[i->to] = true;
                    Q.push(i->to);
                }
            }
        }
    }
}
int map[11][11];
int DP[1<<11][11];
void build(int state, int last, int n) {
    if(state == 0)
        DP[state][last] = map[last][0];
    if(DP[state][last] != none)
        return;
    int i, min = 0xfffffff;
    for(i = 0; i < n; i++) {
        if(state&(1<<i)) {
            build(state^(1<<i), i, n);
            if(min > DP[state^(1<<i)][i]+map[last][i])
                min = DP[state^(1<<i)][i]+map[last][i];
        }
    }
    DP[state][last] = min;
}
void solve(int S[], int s, int n) {
    int i, j, dis[100000];
    for(i = 0; i < s; i++) {
        memset(dis, 127, n<<2);
        SPFA(n, S[i], dis);
        for(j = 0; j < s; j++) {
            map[i][j] = dis[S[j]];
        }
    }
    memset(DP, -1, sizeof(DP));
    build((1<<s)-2, 0, s);
    printf("%d\n", DP[(1<<s)-2][0]);
}
int main() {
    int t, n, m, i;
    int x, y, D, s, S[20];
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        for(i = 0; i < n; i++) {
            link[i].clear();
        }
        Arc arc;
        while(m--) {
            scanf("%d %d %d", &x, &y, &D);
            arc.to = y, arc.v = D;
            link[x].push_back(arc);
            arc.to = x, arc.v = D;
            link[y].push_back(arc);
        }
        scanf("%d", &s);
        for(i = 1; i <= s; i++) {
            scanf("%d", &S[i]);
        }
        S[0] = 0;
        solve(S, s+1, n);
    }
    return 0;
}