HDU 5542 The Battle of Chibi(树状数组优化+dp) - dblank

HDU 5542 The Battle of Chibi(树状数组优化+dp)

The Battle of Chibi

Time Limit: 6000/4000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 809    Accepted Submission(s): 291

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(1100). T test cases follow.

Each test case begins with two numbers N(1N103) and M(1MN), indicating the number of information and number of information Gai Huang will select. Then N numbers in a line, the ith number ai(1ai109) 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
2 3 2 1 2 3 3 2 3 2 1

dp[i][j]表示以第i个数结尾长度为j的上升序列的个数,dp[i][j] = sum(dp[1~x][j-1]),其中x是出现在num[i]前面并且比num[i]小的数,这一个地方用树状数组

优化.,复杂度O(n^2logn).

#include<iostream>  
#include<algorithm>  
#include<cstdio>  
#include<queue>  
#include<map>  
#include<vector>  
#include<cstring>  
using namespace std;  
const int M = 1e9 + 7, maxn = 1010;  
int dp[maxn][maxn], n, m, num[maxn];  
struct node  
{  
    int num, id;  
} s[maxn];  
int cmp(node a, node b)  
{  
    if(a.num != b.num)  
        return a.num < b.num;  
    return a.id > b.id;  
}  
int lowbit(int x)  
{  
    return x & (-x);  
}  
void update(int x, int y, int va)  
{  
    while(x <= n)  
    {  
        dp[x][y] = (dp[x][y] + va) % M;  
        x += lowbit(x);  
    }  
}  
int getsum(int x, int y)  
{  
    int sum = 0;  
    while(x >= 1)  
    {  
        sum = (sum + dp[x][y]) % M;  
        x -= lowbit(x);  
    }  
    return sum;  
}  
int main()  
{  
    int t;  
    cin>>t;  
    for(int cas = 1; cas<=t; cas++)  
    {  
        scanf("%d%d", &n, &m);  
        for(int i = 1; i<=n; i++)  
        {  
            scanf("%d", &s[i].num);  
            s[i].id = i;  
        }  
        sort(s+1, s+n+1,cmp);//离散化  
        memset(dp, 0, sizeof(dp));  
        for(int i = 1; i<=n; i++)  
        {  
            for(int j = 1; j<=m; j++)  
            {  
                if(j == 1)  
                    update(s[i].id, 1, 1);  
                else  
                {  
                    int sum = getsum(s[i].id - 1, j - 1);//求前面长度为j-1,并且满足上升的序列个数和  
                    update(s[i].id, j, sum);//把dp[s[i].id][j]插入到数组里面  
                }  
            }  
        }  
        printf("Case #%d: %d\n", cas, getsum(n, m));  
    }  
    return 0;  
}

 

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