CodeForces 458C Elections

Elections

题意：贿赂选举人使得自己成功获选， 现在求得最小花费是多少。

题解：我们可以发现他是一个凹型函数， 我们用三分去找最小值就好了。

代码：

#include<iostream>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<cstdio>
#define LL long long
#define ULL unsigned LL
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
using namespace std;
const int N = 1e5+5;
vector<int> son[N];
int lt[N];
int cal(int x){
    int need = x - son[0].size();
    int cnt = 0, ret = 0;
    for(int i = 1; i < N; i++){
        for(int j = 0; j < son[i].size(); j++){
            if(son[i].size() - j >= x){
                ret += son[i][j];
                need--;
            }
            else lt[cnt++] = son[i][j];
        }
    }
    if(need <= 0) return ret;
    sort(lt, lt+cnt);
    for(int i = 0; i < need; i++) ret += lt[i];
    return ret;
}
int main(){
    int n, u, v;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++){
        scanf("%d%d",&u,&v);
        son[u].push_back(v);
    }
    for(int i = 1; i < N; i++) sort(son[i].begin(), son[i].end());
    int l = 0, r = n;
    while(l+2 < r){
        int m1 = l+r >> 1;
        int m2 = m1+r >> 1;
        if(cal(m1) > cal(m2)) l = m1;
        else r = m2;
        //printf("%d %d %d %d
",m1, m2, l, r);
    }
    int ans = cal(l);
    for(int i = l+1; i <= r; i++) ans = min(ans,cal(i));
    printf("%d
", ans);
    return 0;
}