强行修炼单调栈系列

POJ2796 神奇的题

呵呵哒 真的是强行修炼受了内伤 我需要冷静 代码如下

  1 #include<cstdio>
  2 #include<map>
  3 //#include<bits/stdc++.h>
  4 #include<vector>
  5 #include<stack>
  6 #include<iostream>
  7 #include<algorithm>
  8 #include<cstring>
  9 #include<cmath>
 10 #include<queue>
 11 #include<cstdlib>
 12 #include<climits>
 13 #define PI acos(-1.0)
 14 #define INF 0x3f3f3f3f
 15 using namespace std;
 16 typedef long long ll;
 17 typedef __int64 int64;
 18 const ll mood=1e9+7;
 19 const int64 Mod=998244353;
 20 const double eps=1e-9;
 21 const int N=2e7+10;
 22 const int MAXN=1e5+5;
 23 inline void rl(ll&num){
 24     num=0;ll f=1;char ch=getchar();
 25     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 26     while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
 27     num*=f;
 28 }
 29 inline void ri(int &num){
 30     num=0;int f=1;char ch=getchar();
 31     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 32     while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
 33     num*=f;
 34 }
 35 int getnum()//相邻的个位整数输入 如想分别保存1234 输入连续的1234 a[i]=getnum();就可以实现
 36 {
 37     char ch=getchar();
 38     while((ch<'0' || ch>'9') && ch!='-')
 39         ch=getchar();
 40     return (ch-'0');
 41 }
 42 inline void out(int x){ if(x<0) {putchar('-');  x*=-1;}if(x>9) out(x/10);    putchar(x%10+'0'); }
 43 struct node{
 44     ll num,p,n;
 45     ll now;
 46 };
 47 ll st[MAXN],t[MAXN];
 48 int main()
 49 {
 50 
 51     int n;
 52     while(scanf("%d",&n)!=EOF)
 53     {
 54         stack<node>s;
 55         node tem;
 56         ll ans=-100,sum=-100;
 57         st[0]=0;
 58         for(int i=1;i<=n;i++)
 59         {
 60             scanf("%lld",t+i);
 61             st[i]=st[i-1]+t[i];
 62         }
 63         tem.num=t[1];tem.p=1;tem.n=tem.now=1;
 64         s.push(tem);
 65         ll x=0,y=0;
 66         for(int i=2;i<=n;i++)
 67         {
 68             node temp;
 69             temp=(node){t[i],1,1,i};
 70             while(!s.empty()&&temp.num<=s.top().num)
 71             {
 72                 tem=s.top();
 73                 s.pop();
 74                 if(!s.empty()) s.top().n+=tem.n;
 75                 temp.p+=tem.p;
 76                 ans=tem.num*(st[tem.now+tem.n-1]-st[tem.now-tem.p]);
 77                 if(ans>sum)
 78                 {
 79                     sum=ans;
 80                     x=tem.now-tem.p+1;
 81                     y=tem.now+tem.n-1;
 82                 }
 83             }
 84             s.push(temp);
 85         }
 86         while(!s.empty())
 87         {
 88             tem=s.top();
 89             s.pop();
 90             if(!s.empty()) s.top().n+=tem.n;
 91             ans=tem.num*(st[tem.now+tem.n-1]-st[tem.now-tem.p]);
 92             if(ans>sum)
 93             {
 94                 sum=ans;
 95                 x=tem.now-tem.p+1;
 96                 y=tem.now+tem.n-1;
 97             }
 98         }
 99         cout<<sum<<endl<<x<<' '<<y<<endl;
100     }
101     return 0;
102 }
神奇的写法

 

 

posted @ 2016-04-19 22:25  zxMrlc  阅读(122)  评论(0)  编辑  收藏  举报