| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 22879 | Accepted: 6279 | |
| Case Time Limit: 1000MS | Special Judge |
Description
Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people's memories about some period of life.A new idea Bill has recently developed assigns a non-negative integer value to each day of human life.
Bill calls this value the emotional value of the day. The greater the emotional value is, the better the daywas. Bill suggests that the value of some period of human life is proportional to the sum of the emotional values of the days in the given period, multiplied by the smallest emotional value of the day in it. This schema reflects that good on average period can be greatly spoiled by one very bad day.
Now Bill is planning to investigate his own life and find the period of his life that had the greatest value. Help him to do so.
Input
The first line of the input contains n - the number of days of Bill's life he is planning to investigate(1 <= n <= 100 000). The rest of the file contains n integer numbers a1, a2, ... an ranging from 0 to 106 - the emotional values of the days. Numbers are separated by spaces and/or line breaks.Output
Print the greatest value of some period of Bill's life in the first line. And on the second line print two numbers l and r such that the period from l-th to r-th day of Bill's life(inclusive) has the greatest possible value. If there are multiple periods with the greatest possible value,then print any one of them.Sample Input
6 3 1 6 4 5 2
Sample Output
60 3 5
题意：给你一个n长度的数组，ans(l,r) 表示 sum l到r的和 * l 到 r子列中的最小值。求最大的ans，输出其和l，r
思路：前缀和预处理。
最开始的思路是用单调列维护（1 ~n）长度的移动框寻找最优解（TLE）
TLE代码：
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
#define inf 0x3f3f3f3f
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
const int maxn = 110000;
using namespace std;
int a[maxn],sum[maxn];
struct node
{
int place,nub;
}que[maxn];
int n;
ll maxx = 0;
ll l = 0;
ll r = 0;
//int que[maxn];
void solve(int k)
{
int tail = 0;
int head = 1;
for(int i = 1;i <= k-1; ++i)
{
while(que[tail].nub>=a[i] && tail>=head)
{
tail--;
}
tail++;
que[tail].nub=a[i];
que[tail].place=i;
}
for(int i=k;i<=n;++i)
{
while(que[tail].nub>=a[i] && tail>=head)
{
tail--;
}
tail++;
que[tail].nub=a[i];
que[tail].place=i;
while(que[tail].place-que[head].place>=k)
{
head++;
}
//printf("%d ",que[head].nub);
ll temp = que[head].nub * (sum[i] - sum[i-k]);
// maxx = max(temp,maxx );
if(temp > maxx)
{
maxx = temp;
l = i - k + 1;
r = i;
}
}
//printf("%I64d\n",maxx );
}
int main()
{
//freopen("C:\\ACM\\input.txt","r",stdin);
while(scanf("%d",&n) != EOF)
{
memset( sum,0 ,sizeof(sum));
maxx = 0;
l = 1;
r = 1;
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)
{
solve(i);
}
//cout<<maxx<<endl<<l<<" "<<r<<endl;
printf("%I64d\n",maxx );
printf("%I64d %I64d\n",l,r);
}
}
正确思路: 遍历每一个a[n]，寻找对于每一个a【n】，求出关于它为最小值的子列的左位置和右位置的大小并且存入数组 l【】 ，r【】中。最后再遍历跑一遍求的最优解
求的左位置和右位置只需要跑两次单调栈就ok了
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
#define inf 0x3f3f3f3f
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
using namespace std;
int a[1100000];
ll sum[1100000];
int l[1100000];
int r[1100000];
int n;
ll maxx = 0;
//int que[maxn];
int main()
{
//freopen("C:\\ACM\\input.txt","r",stdin);
while(scanf("%d",&n) != EOF)
{
memset( sum,0 ,sizeof(sum));
for(int i = 1;i <= n; ++i)
{
scanf("%d",a+i);
sum[i] = sum[i-1] + a[i];
}
stack<int>sta;
for(int i = 1;i <= n; ++i)
{
while(sta.size() && a[i] <= a[sta.top()]) sta.pop();
if(sta.size()) l[i] = sta.top() + 1;
else l[i] = 1;
sta.push(i);
}
while(sta.size()) sta.pop();
for(int i = n;i >= 1; --i)
{
while(sta.size() && a[i] <= a[sta.top()]) sta.pop();
if(sta.size()) r[i] = sta.top() - 1;
else r[i] = n;
sta.push(i);
}
//for(int i = 1;i <= n; ++i) cout<<l[i]<<" "<<r[i]<<endl;
int ansl = 1;
int ansr = 1;
ll ans = 0;
for(int i = 1;i <= n; ++i)
{
ll temp = a[i] * ( sum[r[i]] - sum[l[i]-1] );
if(temp > ans)
{
ans = temp;
ansl = l[i];
ansr = r[i];
}
}
printf("%I64d\n",ans);
printf("%d %d\n",ansl,ansr );
}
}