hdu 5596 bestcode #66 1002

最新推荐文章于 2019-04-09 12:22:12 发布

爱生活爱丙妍

最新推荐文章于 2019-04-09 12:22:12 发布

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分类专栏：思维题

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本文描述了一个有趣的游戏场景，n个游戏角色分为两组进行对决，通过不断增益能量和淘汰对手来比拼生存能力。最终目标是计算经过特定轮次后剩余的角色数量。

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GTW likes gt

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 800 Accepted Submission(s): 291

Problem Description

Long long ago, there were

n adorkable GT. Divided into two groups, they were playing games together, forming a column. The

i−th GT would randomly get a value of ability

bi . At the

i−th second, the

i−th GT would annihilate GTs who are in front of him, whose group differs from his, and whose value of ability is less than his.

In order to make the game more interesting, GTW, the leader of those GTs, would emit energy for

m times, of which the

i−th time of emitting energy is

ci . After the

ci second,

b1,b2,...,bci would all be added 1.

GTW wanted to know how many GTs would survive after the

n−th second.

Input

The first line of the input file contains an integer

T(≤5) , which indicates the number of test cases.

For each test case, there are

n+m+1 lines in the input file.

The first line of each test case contains 2 integers

n and

m , which indicate the number of GTs and the number of emitting energy, respectively.

(1≤n,m≤50000)

In the following

n lines, the

i−th line contains two integers

ai and

bi , which indicate the group of the

i−th GT and his value of ability, respectively.

(0≤ai≤1,1≤bi≤106)

In the following

m lines, the

i−th line contains an integer

ci , which indicates the time of emitting energy for

i−th time.

Output

There should be exactly

T lines in the output file.

The

i−th line should contain exactly an integer, which indicates the number of GTs who survive.

Sample Input

Sample Output

  
   3

   
    
     Hint
    After the first seconds,$b_1=4,b_2=2,b_3=3,b_4=1$
After the second seconds,$b_1=4,b_2=2,b_3=3,b_4=1$
After the third seconds,$b_1=5,b_2=3,b_3=4,b_4=1$,and the second GT is annihilated by the third one.
After the fourth seconds,$b_1=6,b_2=4,b_3=5,b_4=2$
$c_i$ is unordered.

Source

BestCoder Round #66 (div.2)

官方思路：

首先这道题有一个很显然的 $O (n * l o g n)$ 的做法，直接区间加，求区间最大值即可。但是此题还有一个 $O (n)$ 的做法。我们发现 $b_1,b_2,...,b_x$ 都加 $1$ 就相当于 $b_{x+1},b_{x+2},...,b_n$ 都减 $1$ 。然后我们可以倒着做，记一下最大值，如果遇到了修改操作，就把最大值减 $1$ ，然后判断一下这个人会不会被消灭掉，然后再更新一下最大值。

我的代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
struct node
{
    int x;
    int op;
}data[50005];
int main()
{
    int t;
    scanf("%d",&t);
    int c[50005];

    while(t--)
    {//cout<<"1"<<endl;
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&data[i].op,&data[i].x);
        }
        memset(c,0,sizeof(c));
        int x;
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&x);
            c[x+1]++;
        }
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            data[i].x+=m-cnt;
            cnt+=c[i+1];
        }
        int max0=-10000000;
        int max1=-10000000;
        for(int i=n;i>=1;i--)
        {
            if(data[i].op)
            {
                if(data[i].x<max0)
                {
                    n--;
                }
                max1=max(max1,data[i].x);
            }
            else
            {
                if(data[i].x<max1)
                n--;
                max0=max(max0,data[i].x);
            }
        }
        cout<<n<<endl;
    }
    return 0;
}
/*
2
4 4
0 3
1 2
0 3
1 1
1 2 2 4
*/