2015CCPC南阳场 C - The Battle of Chibi

C - The Battle of Chibi

Time Limit:4000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 5542

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   information to be leaked, in happening order. Each of the information was estimated to has   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   information with strict increasing value in happening order. In other words, Gai Huang will not change the order of the  information and just select   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,  .   test cases follow. 

Each test case begins with two numbers   and  , indicating the number of information and number of information Gai Huang will select. Then   numbers in a line, the   number   indicates the value in Cao Cao's opinion of the   information in happening order.

 

Output

For each test case, output one line containing  Case #x: y, where   is the test case number (starting from 1) and   is the ways Gai Huang can select the information. 

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

 

Sample Input

      

       2
3 2
1 2 3
3 2
3 2 1 
      

 

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. 

题意：就是给你n个数，然后找里面有多少个长度为m的严格递增子序列。

思路：这道题我们用dp去做，dp[i][j]表示以第i个数结尾长度为j的严格递增子序列有多少个。显然可以得出递推式：dp[i][j]=∑dp[x][j-1](其中1<=x<i&&a[x]<a[i])。

那么剩下的就是如何求和的问题。如果直接暴力时间复杂度为n^3，显然是不行的。所以这里要用到树状数组优化就能ac了。还要注意一个问题就是，我们用到了树状数组，为了保证肯定能累加，也就是符合上面的a[x]<a[i]，我们要按照a从小到大的顺序dp并且更新树状数组。下面给代码。

#include<iostream>  
#include<stack>  
#include<cstring>  
#include<map>  
#include<string>  
#include<queue>  
#include<algorithm>  
#include<cstdio>  
#include<utility>  
using namespace std;
#define maxn 1005  
#define MOD 1000000007  
int n, m, a[maxn], dp[maxn][maxn], c[maxn][maxn], r[maxn];//c为树状数组，r数组表示a数组的元素从小到大的下标,例如r[i]=j,j为a数组里面第i个小的数的下标  
int lowbit(int x) {
	return x&-x;
}
bool cmp(int x, int y) {
	return a[x] < a[y] || (a[x] == a[y] && x>y);//两个值相等时，坐标后的优先，虽然没有这个判断也能过，但后来队友发现了，加了上去，题目并没说没有相同数据，所以我觉得还是需要加这个判断。
}
void update(int i, int j, int value) {
	while (i <= n) {
		c[i][j] += value;
		c[i][j] %= MOD;
		i += lowbit(i);
	}
}
int sum(int i, int j) {
	int sumnum = 0;
	while (i >= 1) {
		sumnum += c[i][j];
		sumnum %= MOD;
		i -= lowbit(i);
	}
	return sumnum;
}
int main() {
	int t;
	scanf("%d", &t);
	for (int tcase = 1; tcase <= t; tcase++) {
		scanf("%d%d", &n, &m);
		memset(c, 0, sizeof(c));
		for (int i = 1; i <= n; i++) {
			scanf("%d", &a[i]);
		}
		for (int i = 1; i <= n; i++)
			r[i] = i;
		sort(r + 1, r + 1 + n, cmp);
		for (int i = 1; i <= n; i++) {
			int id = r[i];
			dp[id][1] = 1;
			update(id, 1, 1);
			for (int j = 2; j <= m; j++) {
				dp[id][j] = sum(id - 1, j - 1);
				update(id, j, dp[id][j]);
			}
		}
		int ans = 0;
		for (int i = 1; i <= n; i++) {
			ans = (ans + dp[i][m]) % MOD;
		}
		printf("Case #%d: %d\n", tcase, ans);
	}
}