您的位置:首页 > 其它

poj 2481 树状数组

2014-07-16 18:00 176 查看
要用sort,qsort会超时

第一道树状数组,累死了,简直不像写题解


AC代码如下:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

const int MAX_N = 100000;

int q1[MAX_N+5], f1, e1;
int q2[MAX_N+5], f2, e2;
int dp1[MAX_N+5], dp2[MAX_N];
int cnt1[MAX_N+5], cnt2[MAX_N+5];
int num[MAX_N+5];
int N;

int find1( int f, int e, int n ){
int l = f, r = e, mid;
while( l <= r ){
mid = ( l + r ) / 2;
if( n <= q1[mid] ){
r = mid - 1;
}else{
l = mid + 1;
}
}
return r;
}

int calc1( int f, int e, int n ){
int l = f, r = e, mid;
while( l <= r ){
mid = ( l + r ) / 2;
if( q1[mid] > n ){
r = mid - 1;
}else{
l = mid + 1;
}
}
return r - f  + 1;
}

int find2( int f, int e, int n ){
int l = f, r = e, mid;
while( l <= r ){
mid = ( l + r ) / 2;
if( n >= q2[mid] ){
r = mid - 1;
}else{
l = mid + 1;
}
}
return r;
}

int calc2( int f, int e, int n ){
int l = f, r = e, mid;
while( l <= r ){
mid = ( l + r ) / 2;
if( q2[mid] < n ){
r = mid - 1;
}else{
l = mid + 1;
}
}
return r - f  + 1;
}

void DP1(){
dp1[N-1] = 1;
e1 = N - 1;
f1 = N - 1;
cnt1[N-1] = 1;
q1[f1] = num[N-1];
for( int i = N - 2; i >= 0; i-- ){
if( num[i] <= q1[f1] ){
f1--;
q1[f1] = num[i];
dp1[i] = e1 - f1 + 1;
cnt1[i] = calc1( f1, e1, num[i] );
}else{
int pos = find1( f1, e1, num[i] );
q1[pos] = num[i];
dp1[i] = e1 - pos + 1;
cnt1[i] = calc1( pos, e1, num[i] );
}
//       cout << f1 << "***" << e1 << endl;
//      for( int i = f1; i <= e1; i++ ){
//         cout << q1[i] << " ";
//    }
//   cout << endl;
}
}

void DP2(){
dp2[N-1] = 1;
e2 = N - 1;
f2 = N - 1;
cnt2[N-1] = 1;
q2[f2] = num[N-1];
for( int i = N - 2; i >= 0; i-- ){
if( num[i] >= q2[f2] ){
f2--;
q2[f2] = num[i];
dp2[i] = e2 - f2 + 1;
cnt2[i] = calc2( f2, e2, num[i] );
}else{
int pos = find2( f2, e2, num[i] );
q2[pos] = num[i];
dp2[i] = e2 - pos + 1;
cnt2[i] = calc2( pos, e2, num[i] );
}
}
}

int main(){
int T;

scanf( "%d", &T );
while( T-- ){
scanf( "%d", &N );
for( int i = 0; i < N; i++ ){
scanf( "%d", &num[i] );
}
DP1();
DP2();

int ans = 0;
for( int i = N - 1; i >= 0; i-- ){
ans = max( ans, dp1[i] + dp2[i] - min( cnt1[i], cnt2[i] ) );
}

cout << ans << endl;
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: