UVaLive 4731 Cellular Network (期望DP)

题意：手机在蜂窝网络中的定位是一个基本问题，假设蜂窝网络已经得知手机处于c1,c2,,,cn这些区域中的一个，最简单的方法是同时在这些区域中寻找手机，

但这样做很浪费带宽，由于蜂窝网络中可以得知手机在这不同区域中的概率，因此一个折中的办法就是把这些区域分成ｗ组，然后依次访问，求最小的访问区域数的期望值。

析：dp[i][j] 表示第 i 个属于 j 组的期望最小值。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
int dp[maxn][maxn];
int sum[maxn];

int main(){
    int T;   cin >> T;
    while(T--){
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
        sort(a+1, a+n+1, greater<int>());
        sum[0] = 0;
        for(int i = 1; i <= n; ++i)  sum[i] = sum[i-1] + a[i];

        memset(dp, INF, sizeof dp);
        memset(dp[0], 0, sizeof dp[0]);
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= m; ++j){
                for(int k = j-1; k < i; ++k){
                    int t = dp[k][j-1] + (sum[i] - sum[k]) * i;
                    dp[i][j] = min(dp[i][j], t);
                }
            }
        }
        printf("%.4f
", (double)dp[n][m] / sum[n]);
    }
    return 0;
}