Codeforces Round #503 (by SIS, Div. 2)-C. Elections
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枚举每个获胜的可能的票数+按照花费排序
#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #define ll long long using namespace std; struct node { int id,cost; } a[3004]; bool cmp(node a,node b) { return a.cost<b.cost; } int g[3004]; int f[3004]; int main() { int n,m; while(~scanf("%d%d",&n,&m)) { ll sum=999999999999; ll sum2=0; ll ans; memset(f,0,sizeof(f)); memset(g,0,sizeof(g)); for (int i=1; i<=n; i++) { scanf("%d%d",&a[i].id,&a[i].cost); f[a[i].id]++; } sort(a+1,a+1+n,cmp); for (int i=f[1]; i<=n; i++) //枚举赢需要多少票数 { sum2=0; ans=0; for (int j=2; j<=m; j++) { if (f[j]>=i) { g[j]=(f[j]-i+1);//必须降到比我少一票 sum2+=g[j]; } else { g[j]=0; } } if (i-f[1]<sum2)continue;//如果我最终的票数无法赢别人这个状态就是不满足的 sum2=i-f[1]-sum2;//需要从票数比我低的人的那里买多少 for (int j=1; j<=n; j++) { if (g[a[j].id]!=0) { g[a[j].id]--;//比我高的票数减一 ans+=a[j].cost; } else { if (sum2!=0 && a[j].id!=1)//如果我需要在比我的低的人那里买多少 { sum2--; ans+=a[j].cost; } } } sum=min(sum,ans); } cout<<sum<<endl; } return 0; }
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