solution

对于每个糖糖考虑计算他是不是会被消灭。一个组的糖糖会被消灭仅后面存在一个组的糖糖满足，表示前秒总共增加的能量值。移项得。所以把每个糖糖的能量值都变为。然后从后向前枚举的过程中分别统计两个后面的最大值。判断当前点是否会被消灭。

code

/*
* @Author: wxyww
* @Date:   2020-04-20 17:25:13
* @Last Modified time: 2020-04-20 17:37:11
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 1000010;
ll read() {
    ll x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1; c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - '0'; c = getchar();
    }
    return x * f;
}
int mx[2],col[N],a[N],sum[N];
int main() {
    int T = read();
    while(T--) {
        int n = read(),m = read();
        int ans = n;
        for(int i = 1;i <= n;++i) {
            col[i] = read();a[i] = read();
        }
        memset(sum,0,sizeof(sum));
        for(int i = 1;i <= m;++i)
            sum[read()]++;
        for(int i = 1;i <= n;++i) {
            sum[i] += sum[i - 1];
            a[i] -= sum[i - 1];
        }
        mx[0] = mx[1] = -10000000;

        for(int i = n;i >= 1;--i) {
            if(mx[col[i] ^ 1] > a[i]) {
                ans--;
            }
            mx[col[i]] = max(mx[col[i]],a[i]);
        }
        printf("%d\n",ans);
    }

    return 0;
}