CDOJ 1217 The Battle of Chibi

The Battle of Chibi

Time Limit: 6000/4000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

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 and output

Sample Input	Sample Output
2 3 2 1 2 3 3 2 3 2 1	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.

Source

The 2015 China Collegiate Programming Contest

 

解题：求长度为某个值得最长严格上升子序列的个数

数值数组优化dp

dp[i][j]表示以i结尾，长度为j的lis个数，可以得出转移方程是

[dp[i][j] = sum_{k < i,a[k] < a[i]}dp[k][j-1]]

我们c[i][j]表示长度为i的，对应的方案数的树状数组

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1005;
const int mod = 1000000007;
int c[maxn][maxn],a[maxn],b[maxn];
int sum(int i,int j,int ret = 0) {
    while(j > 0) {
        ret = (ret + c[i][j])%mod;
        j -= j&-j;
    }
    return ret;
}
void add(int i,int j,int val) {
    while(j < maxn) {
        c[i][j] = (c[i][j] + val)%mod;
        j += j&-j;
    }
}
int main() {
    int kase,n,m,cs = 1;
    scanf("%d",&kase);
    while(kase--) {
        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; ++i) {
            scanf("%d",a + i);
            b[i-1] = a[i];
        }
        memset(c,0,sizeof c);
        sort(b, b + n);
        int cnt = unique(b,b + n) - b;
        for(int i = 1; i <= n; ++i)
            a[i] = lower_bound(b,b + cnt,a[i]) - b + 1;
        for(int i = 1; i <= n; ++i) {
            add(1,a[i],1);
            for(int j = 2; j <= m; ++j) {
                int t = sum(j-1,a[i]-1);
                if(!t) break;
                add(j,a[i],t);
            }
        }
        printf("Case #%d: %d
",cs++,sum(m,n+2)%mod);
    }
    return 0;
}