【HDOJ 5653】 Bomber Man wants to bomb an Array.（DP）

【HDOJ 5653】 Bomber Man wants to bomb an Array.（DP）

Bomber Man wants to bomb an Array.

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 389    Accepted Submission(s): 117

Problem Description

Given an array and some positions where to plant the bombs, You have to print the Total Maximum Impact.

Each Bomb has some left destruction capability L and some right destruction capability R which means if a bomb is dropped at ith location it will destroy L blocks on the left and R blocks on the right.
Number of Blocks destroyed by a bomb is L+R+1
Total Impact is calculated as product of number of blocks destroyed by each bomb.
If ith bomb destroys Xi blocks then TotalImpact=X1∗X2∗....Xm

Given the bombing locations in the array, print the Maximum Total Impact such that every block of the array is destoryed exactly once(i.e it is effected by only one bomb).

### Rules of Bombing
1. Bomber Man wants to plant a bomb at every bombing location.
2. Bomber Man wants to destroy each block with only once.
3. Bomber Man wants to destroy every block.

 

Input

There are multi test cases denote by a integer T(T≤20) in the first line.

First line two Integers N and M which are the number of locations and number of bombing locations respectivly.
Second line contains M distinct integers specifying the Bombing Locations.

1 <= N <= 2000

1 <= M <= N

 

Output

as Maximum Total Impact can be very large print the floor(1000000 * log2(Maximum Total Impact)).

Hint:
Sample 1:



Sample 2:

 

Sample Input

210 20 910 30 4 8

 

Sample Output

46438565169925

 

Source

BestCoder Round #77 (div.2)

 

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题目大意:有n个块。在其中m块上装有炸药。

要求引爆这些炸药。每个炸药可以由你给定一个左右引爆边界[L,R]表示向左L个块 向右R个块会被摧毁，即爆炸威力为L+R+1(本身所在的块也被摧毁)

设第i个炸药的爆炸威力为Xi

那么总的爆炸威力为X1*X2*X3*X4*...*Xm

问floor(1000000 * log2(Maximum Total Impact)) floor为向下取整函数 Maximum Total Impact为最大爆炸威力和

求log2就是因为成起来太大了。

利用log的性质，可知log(m,(a*b) ) = log(m,a)+log(m,b)

这样用dp[i]表示炸到第i个块可以得到的最大爆炸威力的log

这样可以枚举所有的炸药，对于每个炸药枚举爆炸范围[L,R] 枚举到左右的炸药即可

这样转移方程就是dp[R] = max(dp[R],dp[L-1]+log(R-L+1)/log2)

代码如下:

#include <iostream>#include <cmath>#include <vector>#include <cstdlib>#include <cstdio>#include <cstring>#include <queue>#include <stack>#include <list>#include <algorithm>#include <map>#include <set>#define LL long long#define Pr pair<int,int>#define fread() freopen("in.in","r",stdin)#define fwrite() freopen("out.out","w",stdout)using namespace std;const int INF = 0x3f3f3f3f;const int msz = 10000;const int mod = 1e9+7;const double eps = 1e-8;double dp[2333];int boom[2333];int main(){//fread();//fwrite();int t,n,m;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);boom[0] = 0;boom[m+1] = n+1;for(int i = 1; i <= m; ++i){scanf("%d",&boom[i]);boom[i]++;}sort(boom+1,boom+m+1);memset(dp,0,sizeof(dp));for(int i = 1; i <= m; ++i){for(int l = boom[i-1]+1; l <= boom[i]; ++l){for(int r = boom[i+1]-1; r >= boom[i]; --r){dp[r] = max(dp[r],dp[l-1]+log((r-l+1)*1.0)/log(2.0));}}}LL ans = floor(1e6*dp[n]);printf("%lld\n",ans);}return 0;}