HDOJ 5596 GTW likes gt(巧解)



GTW likes gt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 324    Accepted Submission(s): 113

Problem Description

Long long ago, there weren adorkable GT. Divided into two groups, they were playing games together, forming a column. Thei−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 form 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 integerT(≤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

14 30 31 20 31 1134

 

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.
 

 

题意：有n个GT排成一列，被分成两组，每个GT都会随机拥有bi点能量，到第i秒时第i个GT要消灭掉排在他前面的且与他不是一个组的比他能量低的GT。为了游戏更有意思，GT族的酋长会发功m次，第i次发功在ci秒，第ci秒结束后，  b1，b2，b3....bci都会增加1。 问第n秒后，还有几个GT。

题解：O(n*n)的傻瓜方法果断错了，比赛结束后看了OI爷的代码，学了一发O(n)法。从GT数列的最后端开始考虑，先将酋长在整个过程中给每个GT的能量全加上。然后从数列尾端开始将当前GT与最大值比较，大于最大值就表示能存活下来，然后更新最大值。

代码如下：

#include<cstdio> #include<cstring>#include<algorithm>using namespace std;int a[50010],b[50010],s[50010];int main(){int n,m,i,c,t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=1;i<=n;++i)scanf("%d%d",&a[i],&b[i]);memset(s,0,sizeof(s));for(i=0;i<m;++i){scanf("%d",&c);s[c]++;}for(i=n;i;--i)//越向前增加的能量越多{s[i]+=s[i+1];b[i]+=s[i];}int ans=0;int max1=-1,max2=-1;for(i=n;i;--i){if(!a[i]){if(max2<=b[i])ans++;max1=max(max1,b[i]);}else{if(max1<=b[i])ans++;max2=max(max2,b[i]);}}printf("%d\n",ans);}return 0;}