codeforces 1020 C Elections(枚举+贪心)

题意：

有 n个人，m个党派，第i个人开始想把票投给党派pi，而如果想让他改变他的想法需要花费ci元。你现在是党派1，问你最少花多少钱使得你的党派得票数大于其它任意党派。

n,m<3000

思路：

枚举“除了党派1之外其他党派最多有的票的数量”，贪心一下即可

坑点：

1.爆int

2.inf太小，要设成0x3f3f3f3f3f3f3f3f

3.sort忘了从第一个开始排(被Wa7支配)

#include<iostream>
#include<iomanip> 
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
#include<list>
    
#define fst first
#define sc second
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 
#pragma Gcc optimize(2)

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;


const int maxn = 10000 + 100;
const int maxm = 5e3 + 100;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
int scan(){
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')
        flag=1;
    else if(ch>='0'&&ch<='9')
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}
//int vis[maxn];
struct vt{
    int id;
    int p;
    ll c;
}v[maxn];
int num[maxn];
int cnt[maxn];
int vis[maxn];
/*bool cmp(vt a, vt b){
    if(a.p==1)return false;
    if(b.p==1)return true;
    if(num[a.id] == num[b.id]){
        return a.c < b.c;
    }
    else return num[a.id]<num[b.id];
}*/
bool cmp(vt a, vt b){
    if(a.p==1)return false;
    if(b.p==1)return true;
    return a.c < b.c;
}
int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    //mem(vis, 0);
    mem(num, 0);
    for(int i = 1; i <= n; i++){
        int p, c;
        scanf("%d %lld",&v[i].p, &v[i].c);
        v[i].id = i;
        num[v[i].p]++;
        
    }
    ll ans = 0x3f3f3f3f3f3f3f3f;
    sort(v+1, v+1+n, cmp);
    for(int i = 0; i < n; i++){
        ll tmp = 0;
        mem(vis, 0);
        for(int j = 1; j <= m ;j++){
            cnt[j] = num[j];
        }
        for(int j = 1; j <= n && v[j].p != 1; j++){
            if(cnt[v[j].p] > i){
                cnt[1]++;
                cnt[v[j].p]--;
                tmp += v[j].c;
                vis[j] = 1;
            }
            
        }
        if(cnt[1] <= i){
            for(int j = 1; j <= n; j++ ){
                if(cnt[1]>i)break;
                if(!vis[j]){
                    tmp += v[j].c;
                    cnt[1]++;
                }
            }
        }
        if(cnt[1]>=i)ans = min(tmp, ans);
        //printf("%lld %d\n", tmp, cnt[1]);
    }
    printf("%lld", ans);
    return 0;
}