HDU 5542 The Battle of Chibi

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The Battle of Chibi

Time Limit: 6000/4000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1796 Accepted Submission(s): 641

Problem Description

Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao Cao's army. But all generals and soldiers of Cao Cao were loyal, it's impossible to convince any of them to betray Cao Cao.

So there is only one way left for Yu Zhou, send someone to fake surrender Cao Cao. Gai Huang was selected for this important mission. However, Cao Cao was not easy to believe others, so Gai Huang must leak some important information to Cao Cao before surrendering.

Yu Zhou discussed with Gai Huang and worked out

N information to be leaked, in happening order. Each of the information was estimated to has

ai value in Cao Cao's opinion.

Actually, if you leak information with strict increasing value could accelerate making Cao Cao believe you. So Gai Huang decided to leak exact

M information with strict increasing value in happening order. In other words, Gai Huang will not change the order of the

N information and just select

M of them. Find out how many ways Gai Huang could do this.

Input

The first line of the input gives the number of test cases,

T(1≤100).

T test cases follow.

Each test case begins with two numbers

N(1≤N≤103) and

M(1≤M≤N), indicating the number of information and number of information Gai Huang will select. Then

N numbers in a line, the

ith number

ai(1≤ai≤109) indicates the value in Cao Cao's opinion of the

ith information in happening order.

Output

For each test case, output one line containing Case #x: y, where

x is the test case number (starting from 1) and

y is the ways Gai Huang can select the information.

The result is too large, and you need to output the result mod by

1000000007(109+7).

Sample Input

Sample Output

Case #1: 3
Case #2: 0
Hint
In the first cases, Gai Huang need to leak 2 information out of 3. He could leak any 2 information as all the information value are in increasing order. 
In the second cases, Gai Huang has no choice as selecting any 2 information is not in increasing order.

///  The Battle of Chibi
///给出长度为n的序列，问这个序列中有多少个长度为m的单调递增序列
///题目思路： dp[i][j] 表示到第i个数字，长度为j的单调递增子序列的个数，需要注意的是去第j个数字
///n^3的复杂度会超时，利用树状数组优化第三层循环
#include<bits/stdc++.h>
#define cle(n) memset(n,0,sizeof(n))
using namespace std;
const int maxn =1010;
const int mod= 1000000007;
#define lowbit(x) ((x)&-(x))
int num[maxn],dp[maxn][maxn],b[maxn],n,m,t,cnt=1;

int sum(int x,int y){///计算y长度的子序列，前x项有多少种情况的和
    int ret=0;
    while(x>0){
        ret=(ret+dp[x][y])%mod; x-=lowbit(x);
    }
    return ret;
}
void add(int x,int y,int d){///更新x及以上的长度，能取到的y的之为d
    while(x<=n){
        dp[x][y]=(dp[x][y]+d)%mod;  x+=lowbit(x);
    }
}
int main()
{
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m); cle(dp);
        for(int i=1;i<=n;i++) scanf("%d",&num[i]), b[i]=num[i];
        sort(b+1,b+1+n);
        for(int i=1;i<=n;i++)  num[i]=lower_bound(b+1,b+1+n,num[i])-b;///num[i]储存的是该位置上第几大的元素

        for(int i=1;i<=n;i++){
            for(int j=1;j<=min(i+1,m);j++)
                if(j==1)add(num[i],1,1);
                else int temp=sum(num[i]-1,j-1), add(num[i],j,temp);
        }
        printf("Case #%d: %d\n",cnt++,sum(n,m));
    }
    return 0;
}