hdu 5596__GTW likes gt

GTW likes gt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 727    Accepted Submission(s): 264

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

1 4 3 0 3 1 2 0 3 1 1 1 3 4

 

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)

 

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看了zx大神的代碼，根據思路寫了一遍，結果一直wa，跟ac代碼的區別就在於MAX數組定義爲了局部變量，而全局變量是自動賦初值0的。

後來發現發現訪問的MAX數組的下標可能到n，因此其實如果數據不那麼水的話，除非每次都初始化，否則還是會wa的。代碼如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=50000+10;
struct node
{
    int a,b;
}f[maxn];
int main()
{
    int _,vis[maxn],n,m,c;
    scanf("%d",&_);
    while(_--) {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++) {
            scanf("%d%d",&f[i].a,&f[i].b);
        }
        int MAX[maxn][3];
        memset(MAX,0,sizeof(MAX));
        memset(vis,0,sizeof(vis));
        for(int i=0;i<m;i++) {
            scanf("%d",&c);
            vis[c-1]++;
        }
        int sum=0;
        for(int i=n-1;i>=0;i--) {
            sum+=vis[i];
            f[i].b+=sum;
        }
        int MAX0=-1,MAX1=-1;
        for(int i=n-1;i>=0;i--) {
            if(f[i].a) {
                MAX1=max(MAX1,f[i].b);
                MAX[i][1]=MAX1;
                MAX[i][0]=MAX0;
            }
            else {
                MAX0=max(MAX0,f[i].b);
                MAX[i][1]=MAX1;
                MAX[i][0]=MAX0;
            }
        }
        int ans=0;
        for(int i=0;i<n;i++) {
            if(f[i].a) {
                ans+=f[i].b<MAX[i+1][0];
            }
            else {
                ans+=f[i].b<MAX[i+1][1];
            }
        }
        printf("%d\n",n-ans);
    }
    return 0;
}