[FROM WOJ]#2133 緊急召集

傳送門

【模板複習】

SOL
中位數是在$(i+1)/2$的位置取得，而帶權中位數是在$\sum a[i]>=sum(a[i])/2$的位置取得，與兩點距離無關，只與權值相關
證明：設最優位置爲$t$
則：$\sum_{i=1}^{t-1}a[i]*dis[i,t]+\sum_{i=t+1}^{n}a[i]*dis[i,t]>=\sum_{i=1}^{t}a[i]*dis[i,t+1]+\sum_{i=t+2}^{n}a[i]*dis[i,t+1]$
$\because dis[t,t]=dis[t+1,t+1]=0$
$\therefore \sum_{i=1}^{t}a[i]*dis[i,t]+\sum_{i=t+1}^{n}a[i]*dis[i,t]>=\sum_{i=1}^{t}a[i]*dis[i,t+1]+\sum_{i=t+1}^{n}a[i]*dis[i,t+1]$
$\therefore \sum_{i=1}^{t}a[i]*(dis[i,t]-dis[i,t+1])>=\sum_{i=t+1}^{n}a[i]*(dis[i,t+1]-dis[i,t])$
$\because dis[i,t]-dis[i,t+1](1<=i<=t)=dis[i,t+1]-dis[i,t](t+1<=i<=n)$
$\therefore \sum_{i=1}^{t}a[i]>=\sum_{i=t+1}^{n}a[i]$
得證

代碼：

#include<bits/stdc++.h>
using namespace std;
int num=1,sum,t;
struct node{string c,d;int val;}a[5001];
inline bool cmp(const node&a,const node&b){
	int register l1=a.d.length(),l2=b.d.length();
	if(l1<l2)return 1;
	if(l1>l2)return 0;
	return a.d<b.d;
}
signed main(){
	ios::sync_with_stdio(false);
	while(cin>>a[num].val>>a[num].d>>a[num].c)sum+=a[num].val,num++;num--;
	sort(a+1,a+num+1,cmp);
	for(int register i=1;i<=num;i++){
		t+=a[i].val;
		if(t>=(sum+1)/2)cout<<a[i].c,exit(0);
	}
}