POJ3186+3190+3250

来源：互联网发布：js获取鼠标当前位置编辑：程序博客网时间：2024/06/05 15:45

总结一块发到这里吧

今天竟然放了重题，还好刚做了不久，印象挺深，然后就开始了GG，

题目都有点队列的意思，题目看的是不少，

在中间问有没有人在A c e的时候，我做e已经一个半小时了，若是单纯的计算需要几台机器

倒是简单，就和公司搬桌子问题一样了，不用贪心了，直接转化为查找最大区间重复范围即可

加上这头牛正在使用第几胎机器，我感到很惆怅，另一道动态规划我傻啊傻的模拟，看别人代码

长度才几百，我就有点慌啊，意识到了问题，（就真的只是意识到了，无作为）最后那个搜索直接没敢看

前几天看大牛的博客，动态规划还得刷，如他说的一通百通

我现在七窍通六窍，哎，一窍不通，有点夸张了

先把题目附上吧

FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.

The treats are interesting for many reasons:

The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day, FJ can retrieve one treat from either end of his stash of treats.
Like fine wines and delicious cheeses, the treats improve with age and command greater prices.
The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000).
Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a.

Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally?

The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.

Input

Line 1: A single integer, N

Lines 2..N+1: Line i+1 contains the value of treat v(i)

Output

Line 1: The maximum revenue FJ can achieve by selling the treats

Sample Input

Sample Output

Hint

Explanation of the sample:

Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).

FJ sells the treats (values 1, 3, 1, 5, 2) in the following order of indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.

题意：大体就是给出n个数，然后每次只能从首尾位置上选一个乘以当前出队的个数，求最大值

思路：直接来个模拟求最大值肯定是首尾位置上小的先出队这样必然是最大值，

但是有一种情况难以解决如果首尾位置相等还要继续往内收敛，

那么就没法解决了这时候就有点觉着是动态规划了好吧我写不出来

一直拖到最后看了题解感觉有点一点就通了代码应该可以秒懂

Source Code

/*#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<cmath>using namespace std;int main(){    int i,j,k;    int n;    int a[2001];    int vis[100001];    while(cin>>n)    {        for(i=0;i<n;i++)        {            cin>>a[i];            }        j=0;        k=n-1;        long long ans=0;        int num=1;        memset(vis,0,sizeof(vis));        int flag=1;        for(int m=0;m<n;m++)        {            flag=0;            for(i=0;i<n;i++)            {                //cout<< vis[i]<<" ";                if(vis[i]==0)                {                    flag=1;                    }                }            //cout<<endl;            if(flag==0)break;            if(a[j]<a[k]&&vis[j]==0&&vis[k]==0)            {                vis[j]=1;                ans+=(a[j]*num);                num++;                j++;                }            else if(a[k]<a[j]&&vis[j]==0&&vis[k]==0)            {                vis[k]=1;                ans+=(a[k]*num);                num++;                k--;                }            }            for(i=0;i<n;i++)            {                if(vis[i]==0)                {                    ans+=(a[i]*num);                    break;                    }                }            cout<<ans<<endl;        }    return 0;    }*/#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<cmath>using namespace std;int max(int a,int b){    return a>b?a:b;    }int ans[2001][2001];int main(){    int i,j;    int n;    int a[2002];    cin>>n;    for(i=1;i<=n;i++)        cin>>a[i];    for(i=n;i>0;i--)        for(j=i;j<=n;j++)            ans[i][j]=max(ans[i+1][j]+a[i]*(n+i-j),ans[i][j-1]+a[j]*(n+i-j));    cout<<ans[1][n]<<endl;}

Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows' heads.

Each cow i has a specified height h_i (1 ≤ h_i≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.

Consider this example:

        ==       ==   -   =         Cows facing right -->=   =   == - = = == = = = = =1 2 3 4 5 6

Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow's hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow's hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!

Let c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

Input

Line 1: The number of cows, N.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.

Output

Line 1: A single integer that is the sum of c ₁ through c_N.

Sample Input

610374122

Sample Output

题意：一周之前刚写过这个再写一次吧

向右寻找比当前位置低的碰到高的则更新位置为下一个求一个总和

思路：

首先第一个元素入队，然后找到比他小的入队，如果第三个比第二个大，则更新二号位置，ans里记下下表即可，每次入队的元素逗比前面小，每次加上的num值包含前面所有比他高的数的个数

Source Code

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int map[80001];int main(){    int i,j;    int n,num,x;    long long ans;    while(cin>>n)    {        memset(map,0,sizeof(map));        num=0;        ans=0;        for(i=1;i<=n;i++)        {            cin>>x;            while(num>0&&map[num]<=x)            {                num--;                }            ans+=num;            num++;            map[num]=x;            }        cout<<ans<<endl;        }    return 0;    }

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows.

Help FJ by determining:

The minimum number of stalls required in the barn so that each cow can have her private milking period
An assignment of cows to these stalls over time

Many answers are correct for each test dataset; a program will grade your answer.

Input

Line 1: A single integer, N

Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

Output

Line 1: The minimum number of stalls the barn must have.

Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

Sample Input

51 102 43 65 84 7

Sample Output

Hint

Explanation of the sample:

Here's a graphical schedule for this output:

Time     1  2  3  4  5  6  7  8  9 10Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..Stall 3 .. .. c3>>>>>>>>> .. .. .. ..Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

题意：这个题开始理解错了，输出第一行还是没错的，后面一度理解为当前需要几台机器

还一度怀疑样例出错了提示愣是看不懂细心看了看才领悟

n头牛挤奶，每头牛都有一个时间段，一台机器只能为一头牛挤奶，问你最少需要几台机器

后面分别是是那一台机器榜这头牛挤奶的

看样例因为2 4 4 7这两头牛没有重复的区间

所以两头牛相当于公用一台机器

理解这点就有点絮头了

思路：单纯的最少的机器很好解决

但是机器的标号问题让人很伤脑筋

对于每个新进来的牛若是他开始的时间在前面结束的后面则说明两头牛可以共用

反之则机器的数量上需要+1 并且新来的需要用当前加的这台机器

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
struct node
{
int l;
int e;
int c;
bool friend operator<(node A,node B)
{
if(A.e==B.e)
return A.l>B.l;
return A.e>B.e;
}
}d[60000];
priority_queue<node>q;
int num[50001];
bool cmp(node A,node B)
{
if(A.l==B.l)
return A.e<B.e;
return A.l<B.l;
}
int main()
{
int n;
int i,j;
int ans=1;
cin>>n;
for(i=1;i<=n;i++)
{
scanf("%d %d",&d[i].l,&d[i].e);
d[i].c=i;
}
sort(d+1,d+n+1,cmp);
q.push(d[1]);
num[d[1].c]=1;
for(i=2;i<=n;i++)
{
node temp;
temp=q.top();
int x=temp.e;
if(x<d[i].l)
{
num[d[i].c]=num[temp.c];
q.pop();
}
else
{
ans++;
num[d[i].c]=ans;
}
q.push(d[i]);
}
cout<<ans<<endl;
for(i=1;i<=n;i++)
cout<<num[i]<<endl;
}

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