标签:blog http os io for ar 2014 amp
题目链接:点击打开链接
题意就是求最大面积
枚举每个柱子作为起点
然后二分两边长度。 求个区间最值。
#include<stdio.h> #include<iostream> #include<math.h> using namespace std; #define ll long long #define N 100100 inline bool rd(int &n){ int x = 0, tmp = 1; char c = getchar(); while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar(); if(c == EOF) return false; if(c == '-') c = getchar(), tmp = -1; while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar(); n = x*tmp; return true; } inline void pt(ll n){ if(n < 0) putchar('-'), n = -n; int len = 0,data[20]; while(n) data[len++] = n%10, n /= 10; if(!len) data[len++] = 0; while(len--) putchar(data[len]+48); } int n; int a[N]; int FMin[N][20]; void Init(){ int i,j; for(i=1;i<=n;i++) FMin[i][0]=a[i]; for(i=1;(1<<i)<=n;i++){ //按区间长度递增顺序递推 for(j=1;j+(1<<i)-1<=n;j++){ //区间起点 FMin[j][i]=min(FMin[j][i-1],FMin[j+(1<<(i-1))][i-1]); } } } int Query(int l,int r){ int k=(int)(log(double(r-l+1))/log((double)2)); return min(FMin[l][k],FMin[r-(1<<k)+1][k]); } int main() { while(rd(n), n){ for(int i = 1; i <= n; i++) rd(a[i]); Init(); ll ans = 0; for(int i = 1; i <= n; i++) { if(ans >= (ll)a[i]*n)continue; int l = 1, r = i-1, L = i; while(l<=r) { int mid = (l+r)>>1; if(Query(mid, i-1) >= a[i]) r = mid-1, L = min(L, mid); else l = mid+1; } l = i+1, r = n; int R = i; while(l <= r) { int mid = (l+r)>>1; if(Query(i+1, mid) >= a[i]) l = mid+1, R = max(R, mid); else r = mid-1; } ans = max(ans, (ll)(R-L+1) * a[i]); } pt(ans);putchar('\n'); } return 0; }
POJ 2559 Largest Rectangle in a Histogram RMQ || 单调栈,布布扣,bubuko.com
POJ 2559 Largest Rectangle in a Histogram RMQ || 单调栈
标签:blog http os io for ar 2014 amp
原文地址:http://blog.csdn.net/qq574857122/article/details/38614763