RMQ -- ST算法 Codeforces Round #291 (Div. 2) D. R2D2 and Droid Army

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<utility>
#include<queue>
#include<set>
#include<map>
#include<math.h>
#include<string>
using namespace std;
#define inf 0x3f3f3f3f
#pragma warning(disable:4996)
#pragma comment(linker, "/STACK:102400000,102400000")
#define ll long long
const double eps = 1e-9;

int p[100010][5];
int dp[100010][18][5];
int main()
{
int i, j, k, n, m, K;
cin >> n >> m >> K;
for (i = 0; i < n;i++)
for (j = 0; j < m; j++)
{
scanf("%d",&p[i][j]);
}
for (k = 0; k < m; k++)
{
for (i = 0; i < n; i++)
{
dp[i][0][k] = p[i][k];
}
}
for (k = 0; k < m; k++)
{
for (j = 1; (1 << j) <= n; j++)
{
for (i = 0; i + (1 << j)-1 < n; i++)
dp[i][j][k] = max(dp[i][j-1][k],dp[i+(1<<(j-1))][j-1][k]);
}
}
int l = 0, r = 0;
int mx = 0, tl=-1, tr=-1;
while (1)
{
int t = log((r - l + 1)*1.0) / log(2.0);
int sum = 0;
for (k = 0; k < m; k++)
{
sum += max(dp[l][t][k],dp[r-(1<<t)+1][t][k]);
}
if (sum <= K)
{
if (r - l + 1>mx)
{
mx = r - l + 1;
tl = l, tr = r;
}
if (r != n - 1) r++;
else if (l != n - 1) l++;
else break;
}
else
{
if (l != n - 1) l++;
}
if (r < l) r++;
if (r == n - 1 && l == n - 1) break;
}
if (tl == -1)
{
for (i = 0; i < m;i++)
if (i != m - 1) printf("0 ");
else printf("0\n");
}
else
{
int t = log((tr - tl + 1)*1.0) / log(2.0);
for (k = 0; k < m; k++)
{
int h = max(dp[tl][t][k],dp[tr-(1<<t)+1][t][k]);
if (k == m - 1) printf("%d\n",h);
else printf("%d ",h);
}
}
//system("pause");
}