[CF1019A]Elections

题目大意：有$n$个人，$m$个政党，每个人都想投一个政党，但可以用一定的钱让他选你想让他选的政党。 现在要$1$号政党获胜，获胜的条件是：票数严格大于其他所有政党。求最小代价

题解：暴力枚举其他政党的最多得票，然后把超过的贪心收买，若$1$号政党得票不够，就继续贪心收买，更新答案，复杂度$O(n^2 log_2 n)$（$n=3000,time;limit=2s$卡过）

卡点：1.用$pb\_ds$的$priority\_queueTIL$，换成$std:priority\_queueAC$

C++ Code：

#include <cstdio>
#include <ext/pb_ds/priority_queue.hpp>
#define int long long
#define maxn 3010
using namespace std;
int n, m, ans = 0x3f3f3f3f3f3f3f3f;
int p[maxn], c[maxn];
struct cmp {
	inline bool operator ()(int a, int b) {return a > b;}
};
//__gnu_pbds::priority_queue<int, cmp, __gnu_pbds::binary_heap_tag> q[maxn], Q;
priority_queue<int, std::vector<int>, cmp > q[3050], Q;
void solve(int mid) {
	for (int i = 1; i <= m; i++) while (!q[i].empty()) q[i].pop();
	for (int i = 1; i <= n; i++) q[p[i]].push(c[i]);
	int res = 0, sz = q[1].size();
	for (int i = 2; i <= m; i++) {
		while (q[i].size() > mid) {
			res += q[i].top();
			q[i].pop();
			sz++;
		}
	}
	if (sz <= mid) {
		int delta = mid - sz + 1;
		while (!Q.empty()) Q.pop();
		for (int i = 2; i <= m; i++) {
			int tmp = 0;
			while (!q[i].empty() && tmp < delta) {
				Q.push(q[i].top());
				q[i].pop();
				tmp++;
			}
		}
		while (sz <= mid && !Q.empty()) {
			res += Q.top();
			Q.pop();
			sz++;
		}
	}
	if (res < ans) ans = res;
}
signed main() {
	scanf("%I64d%I64d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%I64d%I64d", &p[i], &c[i]);
	}
	for (int i = n - 1; ~i; i--) solve(i);
	printf("%I64d
", ans);
	return 0;
}