rmq模板
/*
HDU 3183
*/
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
using namespace std;
const int MAXN=1010;
int a[MAXN];
int dp[MAXN][20];void makeRMQIndex(int n,int b[])//形成最小值下標的RMQ
{for(int i=0;i<n;i++)dp[i][0]=i;for(int j=1;(1<<j)<=n;j++)for(int i=0;i+(1<<j)-1<n;i++)dp[i][j]=b[dp[i][j-1]]<=b[dp[i+(1<<(j-1))][j-1]]?dp[i][j-1]:dp[i+(1<<(j-1))][j-1];//這里一定要加等號，就是相等的時候取下標小的
}
int rmqIndex(int s,int v,int b[])
{int k=(int)(log(v-s+1.0)/log(2.0));return b[dp[s][k]]<=b[dp[v-(1<<k)+1][k]]?dp[s][k]:dp[v-(1<<k)+1][k];//加等號，取小標小的
}
char str[MAXN];
int ans[MAXN];
int main()
{//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);int m;while(scanf("%s%d",&str,&m)!=EOF){int n=strlen(str);for(int i=0;i<n;i++)a[i]=str[i]-'0';makeRMQIndex(n,a);int t=0;int temp=0;for(int i=m;i<n;i++)//找n-m個數，每次從[t,i]中找最小的
{t=rmqIndex(t,i,a);ans[temp++]=a[t++];}t=0;while(t<temp&&ans[t]==0)t++;if(t>=temp)printf("0\n");else{for(int i=t;i<temp;i++)printf("%d",ans[i]);printf("\n");}}return 0;
}
?
#include<stdio.h> #include<iostream> #include<math.h> #include<string.h> using namespace std; const int MAXN=50050;int dpmax[MAXN][20]; int dpmin[MAXN][20];void makeMaxRmq(int n,int b[]) {for(int i=0;i<n;i++)dpmax[i][0]=b[i];for(int j=1;(1<<j)<=n;j++)for(int i=0;i+(1<<j)-1<n;i++)dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+(1<<(j-1))][j-1]); } int getMax(int u,int v) {int k=(int)(log(v-u+1.0)/log(2.0));return max(dpmax[u][k],dpmax[v-(1<<k)+1][k]); } void makeMinRmq(int n,int b[]) {for(int i=0;i<n;i++)dpmin[i][0]=b[i];for(int j=1;(1<<j)<=n;j++)for(int i=0;i+(1<<j)-1<n;i++)dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+(1<<(j-1))][j-1]); } int getMin(int u,int v) {int k=(int)(log(v-u+1.0)/log(2.0));return min(dpmin[u][k],dpmin[v-(1<<k)+1][k]); }int a[MAXN]; int main() {int n,Q;int u,v;while(scanf("%d%d",&n,&Q)!=EOF){for(int i=0;i<n;i++)scanf("%d",&a[i]);makeMaxRmq(n,a);makeMinRmq(n,a);while(Q--){scanf("%d%d",&u,&v);u--;v--;int t1=getMax(u,v);int t2=getMin(u,v);printf("%d\n",t1-t2);}}return 0; }?
轉載于:https://www.cnblogs.com/13224ACMer/p/5269861.html
總結
