由于只要找1~x 中的最大值,然后线段树又容易MLE,所以这里可以用树状数组搞。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>
using namespace std;
const int maxn = 1005;
class BIT {
int val[maxn << 2], n;
inline int lowbit(int x) {
return x & (-x);
}
public:
void init(int __n) {
n = __n;
memset(val, 0, sizeof(val));
}
void addv(int pos, int v) {
while(pos <= n) {
val[pos] = max(val[pos], v);
pos += lowbit(pos);
}
}
int GetMax(int pos) {
int ret = 0;
while(pos >= 1) {
ret = max(ret, val[pos]);
pos -= lowbit(pos);
}
return ret;
}
};
BIT bit[maxn];
int a[maxn], b[maxn], f[maxn][maxn][2];
int n, m, num[maxn << 2], numcnt;
inline int GetID(int Val) {
return lower_bound(num, num + numcnt, Val) - num + 1;
}
int main() {
int T; scanf("%d", &T);
for(int kase = 1; kase <= T; kase++) {
memset(f, 0, sizeof(f));
scanf("%d%d", &n, &m);
numcnt = 0;
for(int i = 1; i <= n; i++) {
scanf("%d%d", &a[i], &b[i]);
num[numcnt++] = a[i];
num[numcnt++] = b[i];
}
sort(num, num + numcnt);
numcnt = unique(num, num + numcnt) - num;
for(int i = 1; i <= n; i++) {
a[i] = GetID(a[i]);
b[i] = GetID(b[i]);
}
for(int i = 0; i <= m; i++) bit[i].init(numcnt);
int ans = 0;
for(int i = 1; i <= n; i++) {
for(int j = m; j >= 0; --j) {
int prev_max = bit[j].GetMax(a[i] - 1);
f[i][j][0] = prev_max + 1;
bit[j].addv(a[i], f[i][j][0]);
ans = max(ans, f[i][j][0]);
if(j == 0) continue;
prev_max = bit[j - 1].GetMax(b[i] - 1);
f[i][j][1] = prev_max + 1;
bit[j].addv(b[i], f[i][j][1]);
ans = max(ans, f[i][j][1]);
}
}
printf("%d
", ans);
}
return 0;
}