HDU 5289 Assignment [RMQ区间查询+二分搜索]

题干

hdu 5289走你

题解

/**
*RMQ区间查询+二分搜索
**/
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF = 1e9;
const int MAX = 100010;
int root[MAX];
int mpmax[MAX*2][65];
int mpmin[MAX*2][65];
void RMQ_init(int n){
    int m = log(n)/log(2);
    for(int i=0;i<MAX;i++){
      for(int j=0;j<65;j++){
         mpmax[i][j] = -INF;
         mpmin[i][j] = INF;
      }
    }
    for(int i=1;i<=n;i++){
       mpmax[i][0] = root[i];
       mpmin[i][0] = root[i];
    }
    for(int j=1;j<=m;j++){
       for(int i=1;i<=n;i++){
          int pos = i + (1 << (j-1));
          mpmax[i][j] = max(mpmax[i][j-1],mpmax[pos][j-1]);
          mpmin[i][j] = min(mpmin[i][j-1],mpmin[pos][j-1]);
       }
    }
}
void query(int l, int r, int& rmax, int& rmin){
    int m = log(r-l+1)/log(2) + 1;
    rmax = max(mpmax[l][m-1],mpmax[r - (1<<(m-1)) + 1][m-1]);
    rmin = min(mpmin[l][m-1],mpmin[r - (1<<(m-1)) + 1][m-1]);
}
int main(){
   //freopen("in.txt","r",stdin);
   int t,n;
   long long k,ans;
   scanf("%d",&t);
   while(t--){
      scanf("%d%I64d",&n,&k);
      for(int i = 1; i <= n; i++)scanf("%d",&root[i]);
      RMQ_init(n);
      ans = 0;
      for(int i = 1; i<= n; i++){
          int l = i,//二分的左端点
          r = n, //二分的有端点
          mid = r,//中间点,刚开始mid=(l + r)/2，结果TLE,改成r就AC了，应该是有某个样例专门卡题
          //所以写二分的时候还是得尽量让效率更高
          max, min, kk;
          while(l <= mid && mid <= r){
              query(l, mid, max, min);
              kk = max - min;
              if(kk>=k){
                 r = mid - 1;
                 mid = (l + r) / 2;
              }else{
                 if(mid == r)break;
                 mid = (mid + r) / 2 + (mid + r) % 2;
              }
          }
          ans = ans + r - i + 1;
      }
      printf("%I64d\n",ans);
   }
   return 0;
}