UVALive 4731 Cellular Network

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題解:

也是比較簡單的DP，

貪心不難想到，大的肯定在前面最好。從大到小排序,dp[i][j] 表示前i個數分為j組

dp[i][j] = min( dp[k][j - 1], i * (sum[i] -sum[k]) );

注意精度問題。不要一開始就算。只需要最後除以sum即可

代碼:

#include<bits/stdc++.h>using namespace std;#define pb push_back#define mp make_pair#define se second#define fs first#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define pii pair<int,int>const int INF = 1000000000;const int maxn = 105;const int M = 4000;int a[ maxn ], sum[ maxn ];int dp[ maxn ][ maxn ];int n, m;int main(){    int T;    scanf( "%d", &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 >() );        for( int i = 1; i <= n; i ++ ) sum[ i ] = sum[ i - 1 ] + a[ i ];        for( int i = 1; i <= n; i ++ ) dp[ i ][ 1 ] = i * sum[ i ];        for( int i = 2; i <= m; i ++ )        for( int j = i; j <= n; j ++ )        {            dp[ j ][ i ] = INF;            for( int k = i - 1; k <= j - 1; k ++ )            dp[ j ][ i ] = min( dp[ j ][ i ], dp[ k ][ i - 1 ] + j * ( sum[ j ] - sum[ k ] ) );        }        printf( "%.4lf\n",( double )dp[ n ][ m ] / ( double )sum[ n ] );    }    return 0;}

 

UVALive 4731Cellular Network