C. Arthur and Table
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.
In total the table Arthur bought hasnlegs, the length of thei-th leg isli.
Arthur decided to make the table stable and remove some legs. For each of them Arthur determined numberdi—the amount of energy that he spends to remove thei-th leg.
A table withklegs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with5legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.
Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.
Input
The first line of the input contains integern(1≤n≤105)—the initial number of legs in the table Arthur bought.
The second line of the input contains a sequence ofnintegersli(1≤li≤105), whereliis equal to the length of thei-th leg of the table.
The third line of the input contains a sequence ofnintegersdi(1≤di≤200), wherediis the number of energy units that Arthur spends on removing thei-th leg off the table.
Output
Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.
Sample test(s)
input
21 53 2
output
2
input
32 4 41 1 1
output
0
input
62 2 1 1 3 34 3 5 5 2 1
output
8
题意：
给你一张桌子，有n条腿，告诉每条腿的长度（l）以及砍掉这条腿的花费（d），，当最长腿的数量超过总数量的1/2，那么合法。问如何使得花费最小达到合法情况。
解法：
分别枚举以当前长度为最长的长度，砍去所有长度大于它的。删去长度小于它且花费最少的。
代码：
#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <ctime>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <assert.h>using namespace std;const int MAXN = 100010;int n;int cnt[100010];int val[100010];int cost[210];struct node{int L;int D;}a[MAXN], aa[MAXN];bool cmp1(node a, node b){if (a.L != b.L)return a.L < b.L;return a.D < b.D;}int main(){while (cin >> n){memset(cnt, 0, sizeof(cnt));memset(val, 0, sizeof(val));memset(cost, 0, sizeof(cost));for (int i = 1; i <= n; i++){cin >> a[i].L;cnt[a[i].L]++;}for (int i = 1; i <= n; i++)cin >> a[i].D;sort(a + 1, a + 1 + n, cmp1);int sum = 0;for (int i = n; i >= 1; i–){sum += a[i].D;if (a[i].L != a[i – 1].L){val[a[i – 1].L] = sum;// 比a[i-1].l 长的桌腿的权值和是多少}}int ans = val[a[1].L];cost[a[1].D]++;for (int i = 2; i <= n;i++){if (a[i].L != a[i – 1].L){int tmp = val[a[i].L];int num_k = i – cnt[a[i].L];if (num_k > 0){for (int k = 1; k <= 200; k++){if (cost[k]){if (cost[k] >= num_k){tmp += k * num_k;num_k = 0;break;}else{tmp += k * cost[k];num_k -= cost[k];}}}}ans = min(ans, tmp);}cost[a[i].D]++;}cout << ans << endl;}return 0;}
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世俗的纷扰，生活的琐碎使人精疲力尽，
