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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 977 Accepted Submission(s): 490
这题挺不错的,学到了离线算法。题目要求两点之间最大边的最小值小于某值的个数。其实就是求点对。
集合划分
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
using namespace std;
int n,m,q,set[50010],num[50010],ans[50010];
struct node
{
int x,y,w;
}e[50010];
struct nod
{
int x,id;
}aa[50010];
bool cmp(node a,node b)
{
return a.w<b.w;
}
bool cmp1(nod a,nod b)
{
return a.x<b.x;
}
int find(int x)
{
if(x!=set[x])
set[x]=find(set[x]);
return set[x];
}
int Union(int a,int b)
{
int fx,fy;
fx=find(a),fy=find(b);
if(fx==fy) return 0;
int t=num[fx]*num[fy];
num[fx]+=num[fy];
num[fy]=0;
set[fy]=fx;
return t;
}
int main()
{
while(scanf("%d%d%d",&n,&m,&q)!=EOF)
{
memset(e,0,sizeof(e));
memset(ans,0,sizeof(ans));
for(int i=1;i<=n;i++)
set[i]=i,num[i]=1;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w);
}
sort(e+1,e+1+m,cmp);
for(int i=1;i<=q;i++)
{
scanf("%d",&aa[i].x);
aa[i].id=i;
}
sort(aa+1,aa+1+q,cmp1);
int k=1,cnt=0;
for(int i=1;i<=q;i++)
{
while(k<=m&&e[k].w<=aa[i].x)
{
cnt+=Union(e[k].x,e[k].y);
k++;
}
ans[aa[i].id]=cnt;
}
for(int i=1;i<=q;i++)
printf("%d\n",ans[i]);
}
return 0;
}
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原文地址:http://www.cnblogs.com/a972290869/p/4259468.html