Elections

http://codeforces.com/problemset/problem/457/C

题意：有n支选票，投给ai，如果要这个人把票投给自己，那么需要花费bi，自己的标号为0。求最少的花费，使得自己选票比选票数目最多的那个人还要多。

题解：三分

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int  save[N];
vector<int> g[N];

int cal(int x) {
    int ans = 0, have = g[0].size(), sn = 0;
    for (int i = 1; i <= 100000; i++) {
        int j = 0;
        if (x <= g[i].size()) {
            for (; j < g[i].size() - x + 1; j++) {
                ans += g[i][j];
                have++;
            }
        }
        for (; j < g[i].size(); j++)
            save[sn++] = g[i][j];
    }
    sort(save, save + sn);
    for (int i = 0; i < x - have; i++)
        ans += save[i];
    return ans;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d", &n);
    int a, b;
    for (int i = 0; i < n; i++) {
        scanf("%d%d", &a, &b);
        g[a].push_back(b);
    }
    for (int i = 1; i <= 100000; i++)
        sort(g[i].begin(), g[i].end());
    int l = 1, r = n;
    while (l < r - 2) {
        int midl = (l * 2 + r) / 3;
        int midr = (l + r * 2) / 3;
        if (cal(midl) > cal(midr)) l = midl;
        else r = midr;
    }
    int ans = INF;
    for (int i = l; i <= r; i++)
        ans = min(ans, cal(i));
    printf("%d\n", ans);
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int  save[N];
vector<int> g[N];

int cal(int x) {
    int ans = 0, have = g[0].size(), sn = 0;
    for (int i = 1; i <= 100000; i++) {
        int j = 0;
        if (x <= g[i].size()) {
            for (; j < g[i].size() - x + 1; j++) {
                ans += g[i][j];
                have++;
            }
        }
        for (; j < g[i].size(); j++)
            save[sn++] = g[i][j];
    }
    sort(save, save + sn);
    for (int i = 0; i < x - have; i++)
        ans += save[i];
    return ans;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d", &n);
    int a, b;
    for (int i = 0; i < n; i++) {
        scanf("%d%d", &a, &b);
        g[a].push_back(b);
    }
    for (int i = 1; i <= 100000; i++)
        sort(g[i].begin(), g[i].end());
    int l = 1, r = n;
    while (l < r-1) {
        int midl = (l + r) >> 1;
        int midr = (midl + r) >> 1;
        if (cal(midl) > cal(midr)) l = midl;
        else r = midr;
    }
    printf("%d\n", min(cal(l),cal(r)));
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}