UVA 1456 Cellular Network 贪心+DP

YY过程：概率大的应该先访问吧，嗯，排序一下试试。然后就AC了。。。

我也不知道怎么证明这样贪心是正确。。

至于DP的过程就是水题了，DP[i][j]表示从第i个开始，分成j组的最小期望

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define pb(a) push_back(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\in.txt","r",stdin);
    freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!=' ')return ch;
    }
    return EOF;
}
int n,w;
int u[110];
int dp[110][110];
int f(int k,int g)
{
    if(dp[k][g]>=0)return dp[k][g];
    if(k>n)
    {
        if(g==0)return 0;
        else return INF;
    }
    int minn=INT_MAX;
    int num=0;
    for(int i=k;i<=n&&n-i>=g-1;i++)
    {
        num+=u[i];
        minn=min(minn,f(i+1,g-1)+i*num);
    }
    return dp[k][g]=minn;
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        scanf("%d%d",&n,&w);
        for(int i=1;i<=n;i++)
            scanf("%d",&u[i]);
        sort(u+1,u+1+n,greater<int>());
        memset(dp,-1,sizeof(dp));
        int num=f(1,w);
        int sum=0;
        for(int i=1;i<=n;i++)sum+=u[i];
        printf("%.4lf
",(double)num/sum);
    }
    return 0;
}