旅行商问题(tsp)之分支定界法

http://soj.sysu.edu.cn/show_problem.php?pid=1001&cid=1816

做了一个晚上的题，真是弱爆了...

其实就是深搜最短路，不过加了一个upper bound用来剪枝，因为数据比较小可以过！

深搜还是要熟悉啊！

#include <bits/stdc++.h>

using namespace std;

double graph[25][25];
int vis[25];
double a[25];
double ub;
double ans;


struct point {
    int x;
    int y;
}p[25];

double dis(point a, point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void dfs(int n, int x, double temp, int cnt)
{
//    printf("%.2lf
", temp);
    if(cnt == n-2)
    {
        ub = min(ub, temp+graph[x][n-1]);
        return ;
    }
    vis[x]=1;
    for(int i=0; i<n-1; i++)
    {
        if(graph[x][i] && !vis[i])
        {
            temp += graph[x][i];
            if(temp < ub)
                dfs(n, i, temp, cnt+1);    
            temp -= graph[x][i];
            vis[i]=0;    
                    
        }
    }
    

}


int main()
{
    int t, n, x, y;
    scanf("%d", &t);
    while(t--)
    {
        int cnt=0;
        ub=0x3f3f3f3f;
        double temp=0;
        memset(graph, 0, sizeof(graph));
        memset(vis, 0, sizeof(vis));
        
        scanf("%d", &n);
        for(int i=0; i<n; i++)
            scanf("%d%d", &p[i].x, &p[i].y);

        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                graph[i][j] = dis(p[i], p[j]);    
                        
        vis[0]=1;
        dfs(n,0,temp,0);
        printf("%.2lf
", ub);
    }
    return 0;
}