HDU 5596 GTW likes gt 暴力～

最新推荐文章于 2016-05-16 17:15:07 发布

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GTW likes gt

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 214 Accepted Submission(s): 71

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)

基本思路：仔细看题你可以发现，如果你是从后往前扫的话，其实你可以先直接吧M次所谓的“发功”操作给做了，而且做了之后并不会影响最终结果。

所谓的发功就是对1-Ci 区间进行+1的操作，用差分约束就可以了～

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stack>
#include <queue>
using namespace std;

const int maxx = 50000+10;
int p[maxx];
int kis[maxx];
int vis[maxx];
int n, m;
int T;

int main( ){
    cin>>T;
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i = 0;i <= n+1 ; ++i) kis[i] = 0;
        for(int i = 1;i <= n; ++i){
            scanf("%d%d",&vis[i],&p[i]);
        }
        for(int i = 1;i <= m; ++i){
            int x;
            scanf("%d",&x);
            kis[1]++;
            kis[x+1]--;
        }
        int sum = 0;
        for(int i = 1;i <= n; ++i){
            sum+=kis[i];
            p[i]+=sum;
        }
        int t[2];
        t[0]=t[1]=-1;
        int ans = n;
        for(int i = n;i >= 1; --i){
            if(t[vis[i]]==-1){
                t[vis[i]] = p[i];
            }else if(t[vis[i]] < p[i]){
                t[vis[i]] = p[i];
            }
            if(t[!vis[i]]!=-1 && t[!vis[i]] > p[i]){
                ans--;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}