简单的RMQ，可我怎么写都WA。不明白，找了一个和我相似的贴过了，要赶着去外婆家。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#define eps 1e-5
#define MAXN 255
#define MAXM 111111
#define INF 1000000000
using namespace std;
int mx[MAXN][MAXN][8], mi[MAXN][MAXN][8];
int n, b, q, a, c;
void rmqinit()
{
int l = int(log(double(n)) / log(2.0)) ;
for(int k = 1; k <= n ; k++)
for(int j = 1; j <= l; j++)
for(int i = 1; i + (1 << (j - 1))- 1 <= n; i++)
{
mx[k][i][j] = max(mx[k][i][j - 1], mx[k][i + (1 << (j - 1 ))][j - 1]) ;
mi[k][i][j] = min(mi[k][i][j - 1], mi[k][i + (1 << (j - 1 ))][j - 1]) ;
}
}
int rmqmax(int lx, int ly, int rx, int ry) // lx, ly为左上角的点 rx ry为右下角的点
{
int l = int(log(double(ry - ly + 1)) / log(2.0));
int ret = -1;
for(int k = lx; k <= rx ; k++)
ret = max(ret, max(mx[k][ly][l], mx[k][ry - (1 << l) + 1][l]));
return ret;
}
int rmqmin(int lx, int ly, int rx, int ry) // lx, ly为左上角的点 rx ry为右下角的点
{
int l = int(log(double(ry - ly + 1)) / log(2.0));
int ret = INF;
for(int k = lx; k <= rx ; k++)
ret = min(ret, min(mi[k][ly][l], mi[k][ry - (1 << l) + 1][l]));
return ret;
}
int main()
{
while(scanf("%d%d%d", &n, &b, &q) != EOF)
{
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
{
scanf("%d",&a);
mx[i][j][0] = mi[i][j][0] = a ;
}
rmqinit();
while(q--)
{
scanf("%d%d", &a, &c) ;
int rx = a + b - 1;
if(rx > n) rx = n;
int ry = c + b - 1;
if(ry > n) ry = n;
printf("%d
", rmqmax(a, c, rx, ry) - rmqmin(a, c, rx, ry)) ;
}
}
return 0;
}
MINE：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string.h>
#include <queue>
#include <cmath>
#include <vector>
using namespace std;
int num[255][255];
int f1[255][255][30];
int f2[255][255][30];
int n,b,q;
const int inf=1000000000;
int rmq_max(int p,int i, int j) {
int k = (int)(log(double(j-i+1)) / log(2.0)), t1;
t1 = max(f1[p][i][k], f1[p][j - (1<<k) + 1][k]);
return t1;
}
int rmq_min(int p,int i, int j) {
int k = (int)(log(double(j-i+1)) / log(2.0)), t2;
t2 = min(f2[p][i][k], f2[p][j - (1<<k) + 1][k]);
return t2;
}
int main(){
int l,u;
while(scanf("%d%d%d",&n,&b,&q)!=EOF){
int k = (int) (log((double)n) / log(2.0));
for(int i=0;i<n;i++){
for(int j=0;j<n;j++)
scanf("%d",&num[i][j]);
for(int j = 0; j < n; j++) {
f1[i][j][0] = num[i][j];
f2[i][j][0] = num[i][j];
}
for(int p = 1; p <= k; p++) {
for(int t = 0; t + (1 << p) - 1 < n; t++) {
int m = t + (1 << (p - 1));
f1[i][t][p] = max(f1[i][t][p-1], f1[i][m][p-1]);
f2[i][t][p] = min(f2[i][t][p-1], f2[i][m][p-1]);
}
}
}
int maxn=-1,minn=inf;
for(int i=1;i<=q;i++){
scanf("%d%d",&u,&l);
l--;u--;
for(int p=u;p<(u+b);p++){
maxn=max(maxn,rmq_max(p,l,l+b-1));
minn=min(minn,rmq_min(p,l,l+b-1));
}
printf("%d
",maxn-minn);
}
}
return 0;
}