UVA12983 The Battle of Chibi

Posted 2022-12-27 hs-black

第一眼能看出来是个dp

O($n^3$) 暴力应该很好想 dp[i][j] = $\sum_k=1^i [a[k] < a[i]] *dp[k][j-1]$

发现dp[i][j] 为前面小于它的数长度为j-1的总和, 用树状数组前缀和优化一下搞成$O(n^2log_n)$

先离散化, dp时先枚举位置,再枚举上升序列的长度

树状数组要开二维哦, 打代码的时候发现dp数组可以用树状数组直接代替(小小的空间优化)

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1005;
const int P = 1e9+7;
int read(void) 
    int x = 0;
    char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) 
        x = (x << 3) + (x << 1) + c - '0';
        c = getchar();
    
    return x;

int T, n, m;
int a[N];
struct node
    int pos, val;
    bool operator < (const node &i) const 
        if (val == i.val) return pos < i.pos;
        return val < i.val;
    
p[N];
bool cmp(node i,node j) 
    return i.pos < j.pos;

long long d[N][N];
inline int low(int x) 
    return x & -x;

int sum(int x,int p) 
    long long val = 0;
    for (;x;x -= low(x)) val += d[p][x];
    return val % P;

void add(int x,int k,int p) 
    for (;x <= n; x += low(x)) d[p][x] = (d[p][x] + k) % P;

int main() 
    T = read();
    int cnt = 0;
    while (T--) 
        cnt++;
        n = read(), m = read();
        for (int i = 1;i <= n; i++) p[i] = (node)i,read();
        sort(p + 1,p + n + 1);
        int x = 0, k = 0;
        for (int i = 1;i <= n; i++) 
            if (x == p[i].val) 
                p[i].val = k;
            else 
                x = p[i].val;
                p[i].val = i;
                k = i;
            
        
        sort(p + 1,p + n + 1, cmp);
        memset(d, 0, sizeof(d));
        long long ans = 0;
        for (int i = 1;i <= n; i++) 
            add(p[i].val, 1, 1);
            for (int j = 2;j <= m; j++) 
                int k = sum(p[i].val-1, j-1);
                add(p[i].val, k, j);
                if (j == m) ans = (ans + k) % P;
            
        
        if (m == 1) ans += n;
        printf ("Case #%d: %lld\n", cnt, ans % P);
    
    return 0;