Elections CodeForces - 1020C (贪心)

大意: 有n个选民, m个党派, 第i个选民初始投$p_i$一票, 可以花费$c_i$改变投票, 求最少花费使得第一个党派的票数严格最大

假设最终第一个党派得票数$x$, 枚举$x$, 则对于所有票数$\ge x$的党派, 贪心选择尽量小的c改选第一个党派.若最后票数还不够, 再从没改过的人中选择尽量小的改选即可

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define pb push_back
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

const int N = 4e5+10, INF = 0x3f3f3f3f;
int n, m;
vector<pii> s[N];
pii a[N];
int vis[N];

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,n) {
		int x, y;
		scanf("%d%d", &x, &y);
		s[x].pb({y,i});
		a[i] = {x==1?1e9:y,i};
	}
	sort(a+1, a+1+n);
	REP(i,2,m) sort(s[i].begin(),s[i].end());
	ll ans = 1e18;
	REP(i,1,n) {
		int tot = 0;
		ll c = 0;
		REP(j,2,m) if (s[j].size()>=i) {
			for (int k=0; k<=s[j].size()-i; ++k) {
				++tot, c+=s[j][k].x, vis[s[j][k].y]=1;
			}
		}
		REP(j,1,n) {
			if (s[1].size()+tot<i&&!vis[a[j].y]) {
				++tot, c+=a[j].x;
			}
		}
		if (s[1].size()+tot==i) {
			ans = min(ans, c);
		}
		REP(i,1,n) vis[i]=0;
	}
	printf("%lld\n", ans);
}