hdu 1875 Krustal最小生成树

http://acm.hdu.edu.cn/showproblem.php?pid=1875

题意：求链接所有岛屿的最小生成树。中文题就不说了。

分析：kruskal算法。

用sqrt的时候ce了一次。然后数组开小了re了一次。不应该犯这种低级错误了。反思己过。

View Code

// 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// aply for the memory of the stack//#pragma comment (linker, "/STACK:1024000000,1024000000")//endconst int maxn = 2010;struct node{    int u,v;    double w;}edges[maxn*4];int pa[maxn];int n,m;double x[maxn],y[maxn];double sumweight=0;//初始化void UFset(){    for(int i=0;i<n;i++){        pa[i]=-1;    }}//查找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;    }}int cmp(const void*a,const void*b){    node aa=*(const node*)a;    node bb=*(const node*)b;    if(aa.w>bb.w) return 1;    else return -1;}void Kruskal(){    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+=edges[i].w; num++;            Union(u,v);        }        if(num>=n-1) break;    }}int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        for(int i=0;i<n;i++){            scanf("%lf%lf",&x[i],&y[i]);        }        int mi=0;        for(int i=0;i<n;i++){            for(int j=i+1;j<n;j++){                double d=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));                if(d>=10.0&&d<=1000.0){                    edges[mi].u=i;  edges[mi].v=j;  edges[mi].w=d;                    mi++;                   // printf("edges[mi]=%lf\n",d);                }                else continue;            }        }        m=mi;        qsort(edges,m,sizeof(edges[0]),cmp);        sumweight=0.0;        Kruskal();        if(sumweight==0.0) printf("oh!\n");        else            printf("%.1lf\n",sumweight*100);    }    return 0;}