一組會WA的數據-_-
10 0 0 0 0 0 0 0 0 0 0
問題轉化成:對於每一個數,求出以它爲最小值的最遠左右端點
利用單調棧:維護棧底小,棧頂大的單調棧
每個數入棧的時候記錄左端點(爲棧中前一個數的序列位置+1)
出棧的時候記錄右端點(爲令它出棧時的數的序列位置-1)
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define N 300005
#define LL long long
using namespace std;
inline int wread(){
char c(getchar ());int wans(0),flag(1);
while (c<'0'|| c>'9'){if (c=='-') flag=-1;c=getchar ();}
while (c>='0' && c<='9'){wans=wans*10+c-'0';c=getchar ();}
return wans*=flag;
}
int n;
int sav[N];
struct node{
int zhi,pos;
node()
{}node(int x,int y):zhi(x),pos(y){}
}S[N];
int top;
int ml[N],mr[N];
LL sum[N];
int main (){
n=wread();
for (int i(1);i<=n;++i)
sav[i] = wread();
for (int i(1);i<=n;++i)
sum[i] = sum[i-1]+(LL)sav[i];
S[++top] = node(sav[1],1) ;
ml[1]=1;
int r(1);
while (r!=n){
r++;
if (sav[r]>S[top].zhi) ml[r]=S[top].pos+1,S[++top] = node(sav[r],r);
else if (sav[r]==S[top].zhi) ml[r]=ml[S[top].pos],S[++top] = node(sav[r],r);
else {
while (S[top].zhi>sav[r] && top>=1) mr[S[top].pos]=r-1,top--;
if (!top) S[++top]=node(sav[r],r),ml[r] = 1;
else {
if (S[top].zhi==sav[r]) ml[r] = ml[S[top].pos],S[++top] = node(sav[r],r);
else ml[r] = S[top].pos +1,S[++top] = node(sav[r],r);
}
}
}
while (top!=0) mr[S[top].pos] = n,top--;
LL ans(-1);int pr(0);
for (int i(1);i<=n;++i)
ans < ((LL)sav[i]*( sum[mr[i]] - sum[ml[i]-1] )) ? (pr=i,ans=((LL)sav[i]*( sum[mr[i]] - sum[ml[i]-1] ))) : ans=ans;
printf("%lld\n",ans);
printf("%d %d\n",ml[pr],mr[pr]);
return 0;
}