uva216-枚举-简单题

题意:n个计算机通过电缆连接,怎么连接使用的电缆最少

mmp,死wa不过,memset(vis,0,sizeof(per)),太不小心了

#include <iostream>
#include <memory.h>
#include <stdio.h>
#include<math.h>
using namespace std;

struct Node
{
    int x;
    int y;
    Node()
    {
        x = 0;
        y = 0;
    }
};
const int N = 9;
Node per[N];
int res[N];
int n;
double final = 0x7FFFFFFF;
int vis[N];
double px(Node* n1, Node* n2)
{
    return sqrt(
            0.0 + (n1->x - n2->x) * (n1->x - n2->x)
                    + (n1->y - n2->y) * (n1->y - n2->y)) + 16.00;
}

void dfs(int a[], int cur)
{
    if (cur == n)
    {
        double sum = 0;
        for (int i = 0; i < n - 1; i++)
            sum += px(per + a[i], per + a[i + 1]);
        if (sum < final)
        {
            final = sum;
            memcpy(res, a, sizeof(res));
        }
        return;
    }
    for (int i = 0; i < n; i++)
    {
        if (vis[i])
            continue;
        vis[i] = 1;
        a[cur] = i;
        dfs(a, cur + 1);
        vis[i] = 0;
    }
}
int main()
{
    //freopen("d:\\1.txt", "r", stdin);
    int t = 1;
    while (cin >> n && n)
    {
        memset(per, 0, sizeof(per));
        memset(vis, 0, sizeof(vis));
        memset(res,0,sizeof(res));
        final = 0x7FFFFFFF;
        int s, e;
        for (int i = 0; i < n; i++)
        {
            cin >> s >> e;
            Node p;
            p.x = s;
            p.y = e;
            per[i] = p;
        }
        int a[N];
        dfs(a, 0);
        cout << "**********************************************************"
                << endl;
        cout << "Network #" << t << endl;
        for (int i = 0; i < n - 1; i++)
        {
            double tt = px(per + res[i], per + res[i + 1]);
            printf(
                    "Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n",
                    per[res[i]].x, per[res[i]].y, per[res[i + 1]].x,
                    per[res[i + 1]].y, tt);
        }
        printf("Number of feet of cable required is %.2f.\n", final);
        t++;
    }

    return 0;
}