hdu 1161 kruskal求最小生成树注意建图

题意：求链接所有点的最短路径。

分析：最小生成树，kruskal.嗨。居然wa了2次。受不了啊。

// I'm lanjiangzhou
//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

int n;
const int maxn = 10010;
struct edge{
    int u,v;
    double w;
}edges[maxn];
double sum,maxedge;
int pa[maxn];
int ans[maxn];
double sumweight;
int m;
double x[maxn],y[maxn];

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

void UFset(){
    for(int i=0;i<n;i++){
        pa[i]=-1;
    }
}

int findset(int x){
    int s;
    for(s=x;pa[s]>=0;s=pa[s]);
    while(s!=x){
        int tmp=pa[x];
        pa[x]=s;
        x=tmp;
    }
    return s;
}

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

void Kruskal(){
    int num=0;//已选用边的数目
    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+=edges[i].w;
            num++;
            Union(u,v);
        }
        if(num>=n-1)  break;
    }
}

int main(){
    while(scanf("%d",&n)!=EOF){
        double d;
        int mi=0;
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&x[i],&y[i]);
        }
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;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;
        Kruskal();
        printf("%.2lf\n",sumweight);
    }
    return 0;
}