Uva 216 Getting in Line

 Getting in Line 

题目链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=108&page=show_problem&problem=152

题目比较典型，给你n个点的坐标，找出一条连接所有点的最短的路径，并且求出此时每两个点之间的路径长度，是跟前天校赛最后一题的题目类似，当时听题解时说是二分图匹配，我想这题也可以用二分图匹配吧？但还是用了当初的做法，可以用深度查找（回溯），或者用上排列，我纳闷为什么一开始我会认为是最小生成树呢？汗！！根本是不沾边的事情啊，尽管知道怎么做，但还是调试了一段时间啊

排列做法：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iomanip>
#include<algorithm>
#define MAXN 10
#define MAX_VALUE 21474836
using namespace std;
typedef struct dot{
    int x, y;
}dot;
dot maze[MAXN];
double len[MAXN] , last_len[MAXN];
int list[MAXN], last_list[MAXN];
double cal_dis(int i, int j)
{
    return sqrt((double)((maze[i].x-maze[j].x)*(maze[i].x-maze[j].x)+(maze[i].y-maze[j].y)*(maze[i].y-maze[j].y)));
}
int main()
{

    int T=0;
    int n;
    while(cin>>n && n)
    {
        for(int i=0; i<n; ++i)
        cin>>maze[i].x>>maze[i].y;
        for(int i=0; i<n; ++i) list[i] = i;
        double max = MAX_VALUE;
        do
        {
            double ans = 0;
            len[0] = 0;
            for(int i=1; i<n; ++i)
            {
                len[i] = cal_dis(list[i-1], list[i])+16;
                
                ans += len[i];
            }
            if(max > ans)
            {
                max = ans;
                memcpy(last_list, list, sizeof(int)*n);
                for(int i=1; i<n; ++i) last_len[i] = len[i];
            }
        }while(next_permutation(list, list+n));
        printf("**********************************************************\nNetwork #%d\n", ++T);
        for(int i=1; i<n; ++i)
        {
            printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n", 
            maze[last_list[i-1]].x, maze[last_list[i-1]].y, maze[last_list[i]].x, maze[last_list[i]].y, last_len[i]);
        }
        printf("Number of feet of cable required is %.2lf.\n", max);
    }
    return 0;
}

深度查找（回溯）：

#include<stdio.h>
#include<string.h>
#include<math.h>
#define MAXN 10
double maze[MAXN][MAXN], road[MAXN];
int x[MAXN], y[MAXN];
int visit[MAXN];
int record[MAXN], reco[MAXN];
double final[MAXN];
double max;
double calcu_dis(int x1, int y1, int x2, int y2)
{
    int temp = (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2);
    return sqrt((double)temp);
}

void dfs(int n, int last, int cur)
{
    int i, j;
    double sum = 0;
    if(cur == n)
    {
        for(j=0; j<n; ++j)
        {
            sum += road[j];
        }
        if(sum < max)
        {
            max = sum;
            for(j=0; j<n; ++j)
            final[j] = road[j], reco[j] = record[j];    
        }
        return;
    }
    for(i=0; i<n; ++i)
    {
        if(!visit[i])
        {
            visit[i] = 1;
            record[cur] = i;
            road[cur] = calcu_dis(x[last], y[last], x[i], y[i]);
            dfs(n, i, cur+1);
            visit[i] = 0;
        }
    }
    return;
}

int main()
{

    int i, j, n, T=1;
    while(scanf("%d", &n) != EOF && n)
    {
        max = 21474836;
        for(i=0; i<n; ++i)
        scanf("%d%d", &x[i], &y[i]);
        memset(visit, 0, sizeof(visit));
        for(i=0; i<n; ++i)
        {
            visit[i] = 1;
            record[0] = i;
            road[0] = 0;
            dfs(n, i, 1);
            visit[i] = 0; 
        }
        
        printf("**********************************************************\nNetwork #%d\n", T++);
        for(i=1; i<n; ++i)
        {
            printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n", 
            x[reco[i-1]], y[reco[i-1]], x[reco[i]], y[reco[i]], final[i]+16);
        }
        printf("Number of feet of cable required is %.2lf.\n", max+(n-1)*16);
    }
    return 0;
}