HDU5653 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): 246    Accepted Submission(s): 82

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)

 考虑DP，设dp[i][j]=序列位置i到j能得到的最大的破坏值（是log2之后的值）
dp[i][j]=max(dp[i][j],log2[k-i]+dp[k][j]) i<k<=j 并且保证[i,k-1],[k,j]内都有炸弹。
记忆化搜索一下。
还有就是log2函数速度有点慢，所以要预处理出[1,2000]以内的log2值Log2[]，不然会TLE。

/*****************PID:hdu5653*Auth:Jonariguez*****************事实证明Log2太慢，需要预处理。*/#define lson k*2,l,m#define rson k*2+1,m+1,r#define rep(i,s,e) for(i=(s);i<=(e);i++)#define For(j,s,e) For(j=(s);j<(e);j++)#define sc(x) scanf("%d",&x)#define In(x) scanf("%I64d",&x)#define pf(x) printf("%d",x)#define pfn(x) printf("%d\n",(x))#define Pf(x) printf("%I64d",(x))#define Pfn(x) printf("%I64d\n",(x))#define Pc printf(" ")#define PY puts("YES")#define PN puts("NO")#include <stdio.h>#include <string.h>#include <string>#include <math.h>#include <set>#include <map>#include <stack>#include <queue>#include <vector>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;typedef int Ll;Ll quick_pow(Ll a,Ll b,Ll MOD){a%=MOD;Ll res=1;while(b){if(b&1)res=(res*a)%MOD;b/=2;a=(a*a)%MOD;}return res;}const int maxn=2000+10;const double INF=1000000;int vis[maxn][maxn],p[maxn],sum[maxn];double dp[maxn][maxn],Log2[maxn];double DP(int x,int y){    if(x>=y) return 0;    if(vis[x][y]) return dp[x][y];    vis[x][y]=1;    double &ans=dp[x][y];    ans=0;    if(sum[y]-sum[x-1]==1)        return ans=Log2[y-x+1];    int cnt=0,i=x;    while(p[i]==0) i++;    if(i>y) return ans=-INF;     //区间内无炸弹则返回负无穷    i++;    for(;i<=y;i++){        ans=max(ans,Log2[i-x]+DP(i,y));        if(p[i]) break;          //[x,i-1]内保证只有一个炸弹，所以再次遇到炸弹则停止，下同。    }    i=y;    while(p[i]==0) i--;    i--;    for(;i>=x;i--){        ans=max(ans,Log2[y-i]+DP(x,i));        if(p[i]) break;    }    return ans;}int main(){    int i,j,n,T;    for(i=1;i<=2001;i++)        Log2[i]=log2(i);    scanf("%d",&T);    while(T--){        int m;        sc(n);sc(m);        memset(vis,0,sizeof(vis));        memset(p,0,sizeof(p));        for(i=1;i<=m;i++){            int x;            sc(x);x++;            p[x]=1;        }        sum[0]=0;        for(i=1;i<=2001;i++)            sum[i]=sum[i-1]+p[i];        double res=DP(1,n);        printf("%I64d\n",(LL)floor(1000000.0*res));    }    return 0;}