HDOJ 5596-GTW likes gt【思维题】

GTW likes gt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1133    Accepted Submission(s): 403

Problem Description

Long long ago, there were n 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 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

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.
 

 

Source

BestCoder Round #66 (div.2) 

解题思路：

从后向前查找，并且先用dv数组从后向前把每个人的增量都记录下来，最后遍历一下数组，并且维护0和1的最大值，从后向前原先的加上后来发功加的比最大值小的就会被干掉。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int dv[60000];struct node{int tamp;int gong;}wc[100000];int n,m; int main(){int t;scanf("%d",&t);while(t--){int i,j;scanf("%d%d",&n,&m);for(i=1;i<=n;i++){scanf("%d%d",&wc[i].tamp,&wc[i].gong);dv[i]=0;}dv[i+1]=0;for(i=1;i<=m;i++){int gg;scanf("%d",&gg);dv[gg]++;}for(i=n;i>=1;i--){dv[i]+=dv[i+1];}int mm1,mm2;int ans=0;mm1=-1,mm2=-1;for(i=n;i>=1;i--){if(wc[i].tamp==0){if(wc[i].gong+dv[i]>mm1)mm1=wc[i].gong+dv[i];if(wc[i].gong+dv[i]<mm2){ans++;}}else{if(wc[i].gong+dv[i]>mm2)mm2=wc[i].gong+dv[i];if(wc[i].gong+dv[i]<mm1){ans++;}}}printf("%d\n",n-ans);}return 0;}