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Problem

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.

Mean

找出一个长度为N的数列,长度为M的上升子序列的个数。

Analyse

昨天组队做这道题,其实想的已经差不多了,想到了dp,树状数组,还有离散化。不过都觉得可行性不高,而且这几个知识点熟练度也不高。其实就是这么做,还希望自己在以后的训练中加强难题的训练,不要只做水题。

首先这是动态规划,dp[i][j]表示,长度为i的,以i结尾的,上升子序列的个数,那么就有dp[i][j] = sum(dp[i-1][0~j-1]),由于需要求和,那么需要用树状数组加速,由于j会很大,所以排序后离散化一下。这样总的时间复杂度是n*m*logn.

Code

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1024, mod = 1000000007;
int MAXN, N, n, m, a[maxn], b[maxn];

struct BIT{
    int dat[maxn][maxn];
    void clear() { memset(dat, 0, sizeof(dat)); }
    int sum(int id, int i) {
        int s = 0;
        while (i > 0) {
            s = (s + dat[id][i]) % mod;
            i -= (i & -i);
        }
        return s;
    }
    void add(int id, int i, int x) {
        while (i <= MAXN) {
            dat[id][i] = (dat[id][i] + x) % mod;
            i += (i & -i);
        }
    }
}bit;

int main() {
    int T, kase = 0; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            b[i] = a[i];
        }
        sort(b + 1, b + n + 1);
        N = unique(b + 1, b + n + 1) - (b + 1);
        MAXN = 1<<(32-__builtin_clz(N));
        for(int i = 1; i <= n; ++i)
            a[i] = lower_bound(b + 1, b + N + 1, a[i]) - b;
        bit.clear();
        for(int i = 1; i <= n; ++i) {
            bit.add(1, a[i], 1);
            for(int j = 2; j <= m; ++j) {
                int t = bit.sum(j-1, a[i]-1);
                bit.add(j, a[i], t);
            }
        }
        printf("Case #%d: %d\n", ++kase, bit.sum(m, N));
    }
    return 0;
}

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