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\n", (double)dp[n][m] / sum[n]);

    }

    return 0;

}

巴特西

UVaLive 4731 Cellular Network (期望DP)

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