UVA12983 The Battle of Chibi

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;
}