湖南大学ACM程序设计新生杯大赛 - B Build

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述 

In country  A, some roads are to be built to connect the cities。However, due to limited funds, only some roads can be built.That is to say，if the limit is 100$, only roads whose cost are no more than 100$ can be built. 

Now give you n cities, m roads and the cost  of each road wi (i=1..m). There are q queries, for each query there is a number k, represent the limit, you need to output the maximum number of pairs of cities that can reach each other.  

输入描述:

The first line consist of two inte
4000
gers, n,m, n the number of cities and m the number of roads. The next m lines , each line has three integers a,b,w, represent that you can bulid a road between city a and city b with cost w.
The next line an integer q, the number of querries. The next q lines each an integer k ,the limit of the fund.
n<10000, m < 10000, k < 10000, q<10000;

输出描述:

For each querry ,you should output the anwser.

示例1

输入

3 2
1 2 1
2 3 2
1
2

输出

3

题意：

给你一些边，每个边都有个代价，每次问一个x，问你用代价<=x的边来建树，能有几个点是互相连接的。

POINT：

就是类似最小生成树。先给边排个序，然后依次记录每条边连完之后的答案。

然后二分搜索答案就行了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <math.h>
using namespace std;
#define LL long long
const int maxn = 1e4+55;
const int mod = 2e5+7;
const double eps = 1e-8;
int ans[maxn];
int num[maxn],fa[maxn];
struct node
{
int l,r,w;
}len[maxn];

bool cmd(node a , node b)
{
return a.w<b.w;
}
int find(int x)
{
return fa[x]==x?x:fa[x]=find(fa[x]);
}

int main()
{
int n,m;scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++) fa[i]=i,num[i]=1;
for(int i=1;i<=m;i++){
scanf("%d %d %d",&len[i].l,&len[i].r,&len[i].w);
}
len[0].w=0;
sort(len+1,len+1+m,cmd);
ans[0]=0;
int now=0;
for(int i=1;i<=m;i++){
int x=find(len[i].l);
int y=find(len[i].r);
if(y==x){
}else{
now+=num[y]*num[x];
fa[x]=y;
num[y]+=num[x];
}
ans[i]=now;
}
int q;scanf("%d",&q);
while(q--){
int k;scanf("%d",&k);
int l=0,r=m;
while(l<r){
int mid=(l+r+1)>>1;
if(len[mid].w<=k){
l=mid;
}else{
r=mid-1;
}
}printf("%d\n",ans[l]);

}

}