单调栈1001 POJ2796 Feel Good

来源：互联网发布：流程的再造与优化编辑：程序博客网时间：2024/04/30 18:01

题意:
给一个序列代表某个人的第i天的幸福指数,求一个最幸福的时间段
时间段的幸福指数是该时间段幸福指数的和*该时间段内最低的幸福指数
思路:
求出以第i天为最低的幸福指数的最大区间
单调栈的经典问题
注意一下下边的数据…
InPut:
1
0
OutPut:
0
1 1

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<math.h>#include<queue>#include<stack>#include<string>#include<vector>#include<map>#include<set>using namespace std;#define lowbit(x) (x&(-x))typedef long long LL;const int maxn = 100005;const int inf=(1<<28)-1;stack<int>Stack;int A[maxn],Right[maxn],Left[maxn];LL Sum[maxn];int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;++i)    {        scanf("%d",&A[i]);        Sum[i]=Sum[i-1]+A[i];    }    for(int i=1;i<=n;++i)    {        while(!Stack.empty()&&A[i]<A[Stack.top()])        {            Right[Stack.top()]=i-1;            Stack.pop();        }        Stack.push(i);    }    while(!Stack.empty())     {        Right[Stack.top()]=n;        Stack.pop();    }    for(int i=n;i>=1;--i)    {        while(!Stack.empty()&&A[i]<A[Stack.top()])        {            Left[Stack.top()]=i+1;            Stack.pop();        }        Stack.push(i);    }    while(!Stack.empty())    {        Left[Stack.top()]=1;        Stack.pop();    }    LL Ans=-1;    int LeftPoint=-1,RightPoint=-1;    for(int i=1;i<=n;++i)    if(Ans<(Sum[Right[i]]-Sum[Left[i]-1])*A[i])    {        Ans=(Sum[Right[i]]-Sum[Left[i]-1])*A[i];        LeftPoint=Left[i];        RightPoint=Right[i];    }    printf("%lld\n%d %d\n",Ans,LeftPoint,RightPoint);    return 0;}

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