ZOJ1203 kruskal求最小生成树

题意：给定平面上的n个城市的位置，计算连接这n个城市所需路线长度总和的最小值

分析：直接运用kruskal算法求解。这题关键是建图。好好感受下面建图的思想。一开始我还是想按照以前最短路径那样建图，发现cmp的时候就出问题了。嗨，自己弱爆了。

// I'm the Topcoder
//C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
//C++
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <stack>
#include <string>
#include <list>
#include <queue>
#include <map>
#include <vector>
#include <deque>
#include <set>
using namespace std;

//*************************OUTPUT*************************
#ifdef WIN32
#define INT64 "%I64d"
#define UINT64 "%I64u"
#else
#define INT64 "%lld"
#define UINT64 "%llu"
#endif

//**************************CONSTANT***********************
#define INF 0x3f3f3f3f
#define eps 1e-8
#define PI acos(-1.)
#define PI2 asin (1.);
typedef long long LL;
//typedef __int64 LL;   //codeforces
typedef unsigned int ui;
typedef unsigned long long ui64;
#define MP make_pair
typedef vector<int> VI;
typedef pair<int, int> PII;
#define pb push_back
#define mp make_pair

//***************************SENTENCE************************
#define CL(a,b) memset (a, b, sizeof (a))
#define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b))
#define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c))

//****************************FUNCTION************************
template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); }
template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; }
template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; }

// aply for the memory of the stack
//#pragma comment (linker, "/STACK:1024000000,1024000000")
//end

#define maxn 100+10
#define maxm 5000+10
//边
struct edge{
    int u,v;
    double w;
}edges[maxm];//边的组数
int parent[maxm];//parent[i]为顶点i所在集合对应的树中的根节点
int n,m;
double x[maxm],y[maxm];
double sumweight=0;
//double dis[maxm][maxm];

//double juli(node x,node y){
//    return sqrt((x.u-y.u)*(x.u-y.u)*(x.v-y.v));
//}

//初始化
void UFset(){
    for(int i=0;i<n;i++){
        parent[i]=-1;
    }
}

int findset(int x){
    //return (parent[x]!=x?(parent[x]=findset(x)):x);
    int s;//查找的位置
    for(s=x;parent[s]>=0;s=parent[s]);
    while(s!=x){
        //优化方案：压缩路径，是后续的查找操作加速
        int tmp=parent[x];
        parent[x]=s;
        x=tmp;
    }
    return s;
}

//合并
void Union(int R1,int R2){
    int r1=findset(R1), r2=findset(R2);
    int tmp=parent[r1]+parent[r2];
    if(parent[r1]>parent[r2]){
        parent[r1]=r2;
        parent[r2]=tmp;
    }
    else {
        parent[r2]=r1;
        parent[r1]=tmp;
    }
}

//int cmp(const void*a,const void*b){
//    return *(double)
//}
int cmp(const void*a,const void* b){
    edge aa=*(const edge*)a;
    edge bb=*(const edge*)b;
    if(aa.w>bb.w) return 1;
    else return -1;
}

void Krustral(){
    int num=0;
    int u,v;
    UFset();
    for(int i=0;i<m;i++){
        u=edges[i].u;  v=edges[i].v;
        if(findset(u)!=findset(v)){
            //sumweight+=dis[u][v];
            sumweight+=edges[i].w;
            num++;
            Union(u,v);
        }
        if(num>=n-1)  break;
    }
}

int main(){
    double d;//两点之间的距离
    int kase=1;
    while(scanf("%d",&n)!=EOF){
        if(n==0 ) break;
        for(int i=0;i<n;i++){
            //scanf("%lf%lf",&edges[i].u,&edges[i].v);
            scanf("%lf%lf",&x[i],&y[i]);
        }
        int mi=0;//边的序号
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                //dis[i][j]=juli(edges[i],edges[j]);
                d=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
                edges[mi].u=i;  edges[mi].v=j;  edges[mi].w=d;
                mi++;
;            }
        }
        m=mi;
        qsort(edges,m,sizeof(edges[0]),cmp);
        sumweight=0.0;
        Krustral();
        if(kase>1) printf("\n");
        printf("Case #%d:\n",kase++);
        printf("The minimal distance is: %0.2lf\n",sumweight);
    }
    return 0;
}