CodeForces 458C Elections

Posted 2020-10-26 Schenker

Elections

题意：贿赂选举人使得自己成功获选， 现在求得最小花费是多少。

题解：我们可以发现他是一个凹型函数， 我们用三分去找最小值就好了。

代码：

 1 #include<iostream>
 2 #include<cstring>
 3 #include<string>
 4 #include<queue>
 5 #include<vector>
 6 #include<algorithm>
 7 #include<cmath>
 8 #include<iomanip>
 9 #include<cstdio>
10 #define LL long long
11 #define ULL unsigned LL
12 #define lson l,m,rt<<1
13 #define rson m+1,r,rt<<1|1
14 #define fi first
15 #define se second
16 using namespace std;
17 const int N = 1e5+5;
18 vector<int> son[N];
19 int lt[N];
20 int cal(int x){
21     int need = x - son[0].size();
22     int cnt = 0, ret = 0;
23     for(int i = 1; i < N; i++){
24         for(int j = 0; j < son[i].size(); j++){
25             if(son[i].size() - j >= x){
26                 ret += son[i][j];
27                 need--;
28             }
29             else lt[cnt++] = son[i][j];
30         }
31     }
32     if(need <= 0) return ret;
33     sort(lt, lt+cnt);
34     for(int i = 0; i < need; i++) ret += lt[i];
35     return ret;
36 }
37 int main(){
38     int n, u, v;
39     scanf("%d",&n);
40     for(int i = 1; i <= n; i++){
41         scanf("%d%d",&u,&v);
42         son[u].push_back(v);
43     }
44     for(int i = 1; i < N; i++) sort(son[i].begin(), son[i].end());
45     int l = 0, r = n;
46     while(l+2 < r){
47         int m1 = l+r >> 1;
48         int m2 = m1+r >> 1;
49         if(cal(m1) > cal(m2)) l = m1;
50         else r = m2;
51         //printf("%d %d %d %d\n",m1, m2, l, r);
52     }
53     int ans = cal(l);
54     for(int i = l+1; i <= r; i++) ans = min(ans,cal(i));
55     printf("%d\n", ans);
56     return 0;
57 }