ZOJ1203 Swordfish

一.原题链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=203

二.题目大意：给定平面上N个城市的位置，计算连接这N个城市需要总路线最小长度。

三.思路：没有思路，最小生成树模板题，在某一本图论的书上看到的。注意输出格式就好，除第一个输出外，其他每个输出前面都有一个空行，还有保留2位小数。

四,代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>

using namespace std;

const int MAX_SIZE = 102,
          INF = 1<<30,
          MOD = 1000000007;

struct Edge
{
    int u, v;
    double weight;
};

struct Point
{
    double x, y;
};

Point city[MAX_SIZE];
Edge edges[MAX_SIZE*MAX_SIZE];
int nodeNum, edgeNum, pre[MAX_SIZE];

void FUset()
{
    memset(pre, -1, sizeof(pre));
}

int Find(int x)
{
    int root = x, save;
    while(pre[root] >= 0){
        root = pre[root];
    }

    while(x != root){
        save = pre[x];
        pre[x] = root;
        x = save;
    }

    return root;
}

void Union(int x, int y)
{
    int xRoot = Find(x), yRoot = Find(y);
//此时temp必为负数，负多少表示这棵树有几个节点
    int temp = pre[xRoot] + pre[yRoot];

//把节点少的树合并在节点多的树下面。
    if(pre[xRoot] > pre[yRoot]){
        pre[xRoot] = yRoot;
        pre[yRoot] = temp;
    }
    else{
        pre[yRoot] = xRoot;
        pre[xRoot] = temp;
    }

}

bool cmp(Edge a, Edge b)
{
    return a.weight < b.weight;
}

double Kruskal()
{
    int i, buildEdgeNum = 0, u, v;
    double sumWeight = 0;

    sort(edges, edges + edgeNum, cmp);
    FUset();

    for(i = 0; i < edgeNum; i++){
        u = Find(edges[i].u);
        v = Find(edges[i].v);
        if(u != v){
            sumWeight += edges[i].weight;
            Union(u, v);
        }
        if(buildEdgeNum >= nodeNum - 1)
            break;
    }

    return sumWeight;
}

int main()
{
  //  freopen("in.txt", "r", stdin);
    int i, j, test = 1;
    while(cin>>nodeNum && nodeNum){
        for(i = 0; i < nodeNum; i++)
            cin>>city[i].x>>city[i].y;

        edgeNum = 0;
        for(i = 0; i < nodeNum; i++)
        for(j = i + 1; j < nodeNum; j++){
            edges[edgeNum].u = i;
            edges[edgeNum].v = j;
            edges[edgeNum].weight = sqrt( pow(city[i].x - city[j].x, 2) +
                                          pow(city[i].y -  city[j].y, 2) );
            edgeNum++;
        }
        
        if(test > 1)
            printf("\n");
        
        printf("Case #%d:\nThe minimal distance is: %.2f\n", test++, Kruskal());
    }

}