UVA 216 Getting in Line

大意：给你一些定点，让你以代价最小的边将所有的点连起来。

思路：数据范围很小，可以通过最小生成树或者回溯来解决。

最小生成树去写时不知道哪错了，于是用回溯模拟了一遍，相当于模拟一个数组的全排列。

AC CODE：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>     //INT_MAX,整形范围内的最大整数。
#include <algorithm>
using namespace std;

#define INF 0x3f3f3f3f
const int SIZE = 110;

double d[SIZE];
int vis[SIZE], save[SIZE], ans[SIZE];
int n;
double Min;

struct node
{
    double x, y;
}a[SIZE];

double fun(const node a, const node b)
{
    return sqrt((b.x - a.x) * (b.x - a.x) + (b.y - a.y) * (b.y - a.y));
}

void init()
{
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < n; i++) ans[i] = i;
    Min = INF;
}

void dfs(int cur, double sum)
{
    if(cur == n)
    {
        if(Min > sum)
        {
            Min = sum;
            memcpy(ans, save, sizeof(save));
        }
        return ;
    }
    if(sum >= Min) return ;
    for(int i = 0; i < n; i++) if(!vis[i])
    {
        vis[i] = 1;
        save[cur] = i;
        if(cur == 0)
        {
            dfs(cur+1, 0);
        }
        else
        {
            double dis = fun(a[save[cur]], a[save[cur-1]]);
            dfs(cur+1, sum+dis);
        }
        vis[i] = 0;
    }
}

int main()
{
    int times = 0;
    while(~scanf("%d", &n) && n)
    {
        init();
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf", &a[i].x, &a[i].y);
        }
        dfs(0, 0);
        printf("**********************************************************\n");
        printf("Network #%d\n", ++times);
        for(int i = 1; i < n; i++)
        {
            double dis = fun(a[ans[i]], a[ans[i-1]]) + 16.00;
            printf("Cable requirement to connect (%.lf,%.lf) to (%.lf,%.lf) is %.2lf feet.\n", a[ans[i-1]].x, a[ans[i-1]].y, a[ans[i]].x, a[ans[i]].y, dis);
        }
        double tot = (n-1)*16.00;
        printf("Number of feet of cable required is %.2lf.\n", Min+tot);
    }
}

WA CODE：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>     //INT_MAX,整形范围内的最大整数。
#include <algorithm>
using namespace std;

#define INF 0x3f3f3f3f
const int SIZE = 110;


double w[SIZE][SIZE];
double d[SIZE];
int v[SIZE];
int n;

struct node
{
    double x, y;
}a[SIZE];

double fun(const node a, const node b)
{
    return sqrt((b.x - a.x) * (b.x - a.x) + (b.y - a.y) * (b.y - a.y));
}

double Prim(int src)
{
    int i, j, first = 0;
    double cnt = 0;
    int pre = src;
    for(i = 1; i <= n; i++) d[i] = (i == src)? 0:INF;
    for(i = 1; i <= n; i++)
    {
        int x;
        double m = INF;
        for(int y = 1; y <= n; y++) if(!v[y] && m > d[y]) m = d[x=y];
        v[x] = 1;
        cnt += m;
        if(first)
        {
            printf("Cable requirement to connect (%.lf,%.lf) to (%.lf,%.lf) is %.2lf feet.\n", a[pre].x, a[pre].y, a[x].x, a[x].y, m+16);
        }
        for(int y = 1; y <= n; y++) d[y] = min(d[y], w[x][y]);
        pre = x;
        first = 1;
    }
    return cnt;
}

void init()
{
    memset(v, 0, sizeof(v));
    memset(d, 0, sizeof(d));
    memset(a, 0, sizeof(a));
    for(int i = 1; i <= SIZE; i++)
        for(int j = 1; j <= SIZE; j++)
            w[i][j] = w[j][i] = INF;
}



int main()
{
    int i, j;
    int times = 0;
    while(~scanf("%d", &n) && n)
    {
        init();
        for(i = 1; i <= n; i++)
        {
            scanf("%lf%lf", &a[i].x, &a[i].y);
        }
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j <= n; j++)
            {
                if(i == j) continue;
                w[i][j] = w[j][i] = fun(a[i], a[j]);
            }
        }
        printf("**********************************************************\n");
        printf("Network #%d\n", ++times);
        double ans = Prim(1);
        double tot = (n-1)*16.00;
        printf("Number of feet of cable required is %.2lf.\n", ans+tot);
    }
    return 0;
}