HDU 5653 Bomber Man wants to bomb an Array. DP

Bomber Man wants to bomb an Array.

[align=left][b]Problem Description[/b][/align]
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.

[align=left][b]Input[/b][/align]
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

[align=left][b]Output[/b][/align]
as Maximum Total Impact can be very large print the floor(1000000 * log2(Maximum Total Impact)).

[b]Hint:[/b]
Sample 1:



Sample 2:



[align=left][b]Sample Input[/b][/align]

2
10 2
0 9
10 3
0 4 8

[align=left][b]Sample Output[/b][/align]

4643856
5169925

[b]题意：[/b]

给一个长度为 NN 的一维格子和一些炸弹的位置，请你计算 “最大总破坏指数”。

每个炸弹都有向左和向右的破坏力，如果一个炸弹向左和向右的破坏力分别为 LL 和 RR,
那么该炸弹将炸毁 L + R + 1L+R+1 个格子（左边LL个，炸弹所在格子，右边RR个）。
破坏指数的计算方式为：所有炸弹炸毁的格子数的乘积。假设第 ii 个炸弹炸毁了 X_iX​i​​个格子，
那么总破坏指数就是 X_1 * X_2 * .... X_mX​1​​∗X​2​​∗....X​m​​。

现在告诉你每个炸弹的位置，你需要计算 最大的总破坏指数，注意：每个格子最多只允许被炸一次。

[b]题解：[/b]

　DP

　设定f[i][j] 表示前i个炸弹到j的最大破坏力

　我们在得知f[i][k]下 枚举当前炸弹左右范围更新答案就好了

　　

#pragma comment(linker, "/STACK:10240000,10240000")
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
const int  N = 2e3+10 , M = 1e6+20;
using namespace std;
typedef long long ll;
int T,n,m;
double f

;
int a
,x;
int main() {
scanf("%d",&T);
while(T--) {
scanf("%d%d",&m,&n);
for(int i=1;i<=n;i++) scanf("%d",&x), a[i] = x + 1;
sort(a+1,a+n+1);
memset(f,0,sizeof(f));
f[0][0] = 0.0;
a[0] = 0, a[n+1] = m+1;
for(int i=1;i<=n;i++) {
for(int j=a[i];j<a[i+1];j++) {
for(int k=a[i-1]+1;k<=a[i];k++) {
f[i][j] = max(f[i][j], f[i-1][k-1] + log(j-k+1)/ log(2));
}
}
}
printf("%lld\n",(ll)floor(1000000*f
[m]));
}
return 0;
}