旅行商问题

NPC问题，不存在多项式时间的算法，但是在算法中可以做剪枝优化:

1. 第一次写的算法，缺少剪枝：  当前路径 >= 之前算出的最短路径， 则当前路径不在继续遍历

#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

int N;
int map[12][12];
int flag[12];

int solve(int start) {
    int i,val, all_visited = 1, min = 999999999;

    for (i = 2; i <= N; i++) {
        if (flag[i] != 0 || map[start][i] == 0) {
            continue;
        }
        all_visited = 0;
        flag[i] = 1; //visited
        val = map[start][i] + solve(i);
        if (val < min) {
            min = val;
        }
        flag[i] = 0;
    }
    if (all_visited) {
        return map[start][1];
    } else {
        return min;
    }
}

int main(int argc, char** argv)
{
    int tc, T, i, j;
    // freopen("input.txt", "r", stdin);
    cin >> T;
    for(tc = 0; tc < T; tc++)
    {
        cin>>N;
        for (i = 1; i <= N; i++) {
            for (j = 1; j <= N; j++) {
                cin>>map[i][j];
            }
        }

        memset(flag, 0x00, 12*sizeof(int));
        printf("%d
", solve(1));
    }

    return 0;
}

2. 优化算法，路径判断剪枝

#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

int N, cost, min_cost;
int map[13][13];
int flag[13];
int sum;

void solve(int start) {
    int i;

    if (sum >= N) {
        if (map[start][1]!=0 && cost + map[start][1] < min_cost) {
            min_cost = cost + map[start][1];
        }
        return;
    }

    for (i = 2; i <= N; i++) {
        if (flag[i] != 0 || map[start][i] == 0 || cost + map[start][i] >= min_cost) {
            continue;
        }
        flag[i] = 1; //visited
        cost += map[start][i];
        sum++;
        solve(i);
        sum--;
        flag[i] = 0;
        cost -= map[start][i];
    }
}

int main(int argc, char** argv)
{
    int tc, T, i, j;
    //freopen("input.txt", "r", stdin);
    cin >> T;
    for(tc = 0; tc < T; tc++)
    {
        cin>>N;
        for (i = 1; i <= N; i++) {
            for (j = 1; j <= N; j++) {
                    cin>>map[i][j];
            }
        }

        memset(flag, 0x00, 13*sizeof(int));
        cost = 0; min_cost = 99999999999;
        sum = 1;
        solve(1);
        cout<<min_cost<<endl;
    }

    return 0;
}

注意红色代码，不要遗漏