枚举2~m的最大票数，对于枚举的每个票数i，可以用优先队列以最小的代价构造出1号的胜局，然后在所有最小代价中取最小值即可
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 3010;
const ll INF = 1e16;
vector<int> v[maxn];
vector<ll> sum[maxn];
int pos[maxn];
struct Node
{
int val, id;
bool operator < (const Node &t) const
{
return val > t.val;
}
} node;
int main()
{
int n, m;
scanf("%d%d", &n, &m);
int h1 = 0, maxh = 0;
for (int i = 1; i <= n; i++)
{
int x, y;
scanf("%d%d", &x, &y);
v[x].push_back(y);
if (x == 1) h1++;
}
for (int i = 2; i <= m; i++)
{
sort(v[i].begin(), v[i].end());
ll t = 0;
for (int j = 0; j < v[i].size(); j++)
{
t += v[i][j];
sum[i].push_back(t);
}
maxh = max(maxh, (int)v[i].size());
}
if (h1 > maxh)
{
printf("0\n");
return 0;
}
ll ans = INF;
for (int i = h1; i <= maxh; i++)
{
for (int j = 2; j <= m; j++) pos[j] = 0;
int h = h1;
ll cost = 0;
for (int j = 2; j <= m; j++)
{
if (v[j].size() > i)
{
pos[j] = v[j].size()-i;
h += (v[j].size()-i);
cost += sum[j][pos[j]-1];
}
}
if (h >= i + 1)
{
ans = min(ans, cost);
continue;
}
priority_queue<Node> que;
for (int j = 2; j <= m; j++) {
if (pos[j] < v[j].size()) que.push(Node{v[j][pos[j]], j});
}
int need = i+1-h;
while (need)
{
if (que.empty()) break;
node = que.top();
que.pop();
cost += node.val;
int p = node.id;
pos[p]++;
if (pos[p] < v[p].size()) que.push(Node{v[p][pos[p]], p});
need--;
}
if (need == 0) ans = min(ans, cost);
}
printf("%I64d\n", ans);
return 0;
}
