RMQ模板

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 50005
#define MAXN 100005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,m,ans,cnt,tot,flag;
int len,st,ed;   // 直徑 起點 終點
bool vis[maxn];
int head[maxn],val[maxn],dist[maxn][2];
int f[maxn][20],g[maxn][20],lg[maxn];  //第二維為二進位
struct Node
{
int v,w,next;
} edge[MAXN];

void addedge(int u,int v,int w)
{
cnt++;
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=head[u];
head[u]=cnt;
}
void dfs(int u,int fa,int sum){
if(sum>len){
len=sum;
st=u;
}
int i,j,v;
for(i=head[u];i;i=edge[i].next){
v=edge[i].v;
if(v!=fa){
dfs(v,u,sum+edge[i].w);
}
}
}
void dfs1(int u,int fa,int k,int sum){
dist[u][k]=sum;
int i,j,v;
for(i=head[u];i;i=edge[i].next){
v=edge[i].v;
if(v!=fa)
dfs1(v,u,k,sum+edge[i].w);
}
}
void init_rmq()              // 預處理  O(n*log(n))
{
int i,j;
for(i=1;i<=n;i++)
{
f[i][0]=g[i][0]=val[i];
}
for(j=1;(1<<j)<=n;j++)
{
for(i=1;i+j-1<=n;i++)
{
if((i+(1<<(j-1))<=n))
{
f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);  //避免越界
g[i][j]=min(g[i][j-1],g[i+(1<<(j-1))][j-1]);
}
else
{
f[i][j]=f[i][j-1];
g[i][j]=g[i][j-1];
}
}
}
}
int query_rmq(int l,int r)
{
int k=lg[r-l+1];
return
max(f[l][k],f[r-(1<<k)+1][k])-min(g[l][k],g[r-(1<<k)+1][k]);
}
int main()
{
int i,j,t;
lg[1]=0;
for(i=2;i<=50000;i++)
{
lg[i]=lg[i>>1]+1;
}
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0) break ;
cnt=0;
memset(head,0,sizeof(head));
int u,v,w;
for(i=1;i<n;i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
len=0;
dfs(1,0,0);
ed=st;
len=0;
dfs(st,0,0);

dfs1(st,0,0,0);
dfs1(ed,0,1,0);

for(i=1;i<=n;i++)
{
val[i]=max(dist[i][0],dist[i][1]);
}
init_rmq();
int q,le,ri,mid,res,id;
while(m--){
scanf("%d",&q);
ans=id=1;
for(i=1;i<=n;i++)
{
while(id<=i){
res=query_rmq(id,i);
if(res>q)
id++;
else break ;
}
ans=max(ans,i-id+1);
}
printf("%d\n",ans);
}
}
return 0;
}

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

int val[255][255];
int mm[255];
int dpmin[255][255][8][8];//最小值
int dpmax[255][255][8][8];//最大值

void initRMQ(int n,int m)
{
for(int i = 1;i <= n;i++)
for(int j = 1;j <= m;j++)
dpmin[i][j][0][0] = dpmax[i][j][0][0] = val[i][j];
for(int ii = 0; ii <= mm
; ii++)
for(int jj = 0; jj <= mm[m]; jj++)
if(ii+jj)
for(int i = 1; i + (1<<ii) - 1 <= n; i++)
for(int j = 1; j + (1<<jj) - 1 <= m; j++)
{
if(ii)
{
dpmin[i][j][ii][jj] = min(dpmin[i][j][ii-1][jj],dpmin[i+(1<<(ii-1))][j][ii-1][jj]);
dpmax[i][j][ii][jj] = max(dpmax[i][j][ii-1][jj],dpmax[i+(1<<(ii-1))][j][ii-1][jj]);
}
else
{
dpmin[i][j][ii][jj] = min(dpmin[i][j][ii][jj-1],dpmin[i][j+(1<<(jj-1))][ii][jj-1]);
dpmax[i][j][ii][jj] = max(dpmax[i][j][ii][jj-1],dpmax[i][j+(1<<(jj-1))][ii][jj-1]);
}
}
}
//查询矩形的最大值
int rmq1(int x1,int y1,int x2,int y2)
{
int k1 = mm[x2-x1+1];
int k2 = mm[y2-y1+1];
x2 = x2 - (1<<k1) + 1;
y2 = y2 - (1<<k2) + 1;
return max(max(dpmax[x1][y1][k1][k2],dpmax[x1][y2][k1][k2]),max(dpmax[x2][y1][k1][k2],dpmax[x2][y2][k1][k2]));
}
//查询矩形的最小值
int rmq2(int x1,int y1,int x2,int y2)
{
int k1 = mm[x2-x1+1];
int k2 = mm[y2-y1+1];
x2 = x2 - (1<<k1) + 1;
y2 = y2 - (1<<k2) + 1;
return min(min(dpmin[x1][y1][k1][k2],dpmin[x1][y2][k1][k2]),min(dpmin[x2][y1][k1][k2],dpmin[x2][y2][k1][k2]));
}

int main()
{
mm[0] = -1;
for(int i = 1;i <= 500;i++)
mm[i] = ((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
int N,B,K;
for(int i=1;i<=250;i++){
if(i%5==0)
printf("\n");
printf("%10d",mm[i]);
}
while(scanf("%d%d%d",&N,&B,&K)==3)
{
for(int i = 1;i <= N;i++)
for(int j = 1;j <= N;j++)
scanf("%d",&val[i][j]);
initRMQ(N,N);
int x,y;
while(K--)
{
scanf("%d%d",&x,&y);
printf("%d\n",rmq1(x,y,x+B-1,y+B-1)-rmq2(x,y,x+B-1,y+B-1));
}
}
return 0;
}

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

const int MAXN = 100010;

int dp[MAXN][20];
int mm[MAXN];

void initRMQ(int n,int b[])
{
mm[0] = -1;
for(int i = 1;i <= n;i++)
{
mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];
dp[i][0] = b[i];
}
for(int j = 1;j <= mm
;j++)
for(int i = 1;i + (1<<j) - 1 <= n;i++)
dp[i][j] = max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int rmq(int x,int y)
{
int k = mm[y-x+1];
return max(dp[x][k],dp[y-(1<<k)+1][k]);
}

int a[MAXN];
int hash[MAXN];
int L[MAXN],R[MAXN];
int b[MAXN];

int main()
{
int n,m;
while(scanf("%d",&n) == 1 && n)
{
scanf("%d",&m);
for(int i = 1;i <= n;i++)
scanf("%d",&a[i]);

int k = 1;
int l = 1;
memset(b,0,sizeof(b));
for(int i = 1;i <= n;i++)
{
if(i > 1 && a[i] != a[i-1])
{
for(int j = l;j < i;j++)
{
L[j] = l;
R[j] = i-1;
}
b[k] = i - l;
l = i;
k ++;
}
hash[i] = k;
}
for(int j = l;j <= n;j++)
{
L[j] = l;
R[j] = n;
}
b[k] = n-l+1;

initRMQ(k,b);
int u,v;
while(m--)
{
scanf("%d%d",&u,&v);
int t1 = hash[u];
int t2 = hash[v];
if(t1 == t2)
{
printf("%d\n",v-u+1);
continue;
}
int ans = max(R[u]-u+1,v-L[v]+1);
t1++;
t2--;
if(t1 <= t2)ans = max(ans,rmq(t1,t2));
printf("%d\n",ans);
}
}
return 0;
}