UVALive 6924Load Balancing（dp）

题意:

N≤104个数的序列,每个数∈[0,160],现求一个四等分点,使得落在每个区间的频数与浮点平均数的差值和最小

分析:

一开始想着贪心,然后又想暴力枚举,直到最后发现了数是可以重复的,−−160的范围,1W个数。。
然后就是显然的dp了,维护下数的个数的前缀和
考虑状态dp[i][j]:=[0,i]范围的数划分成j部分的最小差值和
然后暴力转移就可以了,记录下转移的过程为了打印方便
嘛,a,b,c是不能相等的,所以转移的时候是<

代码:

////  Created by TaoSama on 2015-12-09//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n;double dp[165][5];int pre[165][5], cnt[165], sum[165];void printDp(int i, int j) {    printf("dp[%d][%d] = %.5f\n", i, j, dp[i][j]);}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d", &n);        memset(cnt, 0, sizeof cnt);        for(int i = 1; i <= n; ++i) {            int x; scanf("%d", &x);            ++cnt[x];        }        sum[0] = cnt[0];        for(int i = 1; i <= 160; ++i) sum[i] = sum[i - 1] + cnt[i];        double avg = n / 4.0;        for(int i = 0; i <= 160; ++i) {            dp[i][1] = abs(sum[i] - avg);            for(int j = 2; j <= 4; ++j) {                dp[i][j] = INF;                for(int k = 0; k < i; ++k) {                    double dif = abs(sum[i] - sum[k] - avg);                    if(dp[i][j] > dp[k][j - 1] + dif) {                        dp[i][j] = dp[k][j - 1] + dif;                        pre[i][j] = k;                    }                }            }        }        vector<int> ans;        int cur = 160, k = 4;        while(ans.size() != 3) {            ans.push_back(pre[cur][k]);            cur = pre[cur][k];            --k;        }        printf("Case %d: %d %d %d\n", ++kase, ans[2], ans[1], ans[0]);    }    return 0;}