Day3-G

Today the company has m tasks to complete. The ith task need xi minutes to complete. Meanwhile, this task has a difficulty level yi. The machine whose level below this task’s level yi cannot complete this task. If the company completes this task, they will get (500*xi+2*yi) dollars. 
The company has n machines. Each machine has a maximum working time and a level. If the time for the task is more than the maximum working time of the machine, the machine can not complete this task. Each machine can only complete a task one day. Each task can only be completed by one machine. 
The company hopes to maximize the number of the tasks which they can complete today. If there are multiple solutions, they hopes to make the money maximum.

InputThe input contains several test cases. 
The first line contains two integers N and M. N is the number of the machines.M is the number of tasks(1 < =N <= 100000,1<=M<=100000). 
The following N lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the maximum time the machine can work.yi is the level of the machine. 
The following M lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the time we need to complete the task.yi is the level of the task.OutputFor each test case, output two integers, the maximum number of the tasks which the company can complete today and the money they will get.Sample Input

1 2
100 3
100 2
100 1

Sample Output

1 50004

思路：本题和H(Alice and Bob)一样，可看作偏序，x的权重大于y，将machine的x大于等于task的y值入队，每次入队最小的大于task的y值的，这样就能保证后面的y可以更多的匹配，
而x在入队时已经得到保证，代码如下：

typedef long long LL;

const int maxm = 100010;

struct Node {
    int x, y;

    bool operator<(const Node &a)const {
        return x > a.x || (x == a.x && y > a.y);
    }
} machine[maxm], task[maxm];

int n, t, tmp[105];

int main() {
    while(scanf("%d%d", &n, &t) == 2) {
        memset(tmp, 0, sizeof(tmp));
        for (int i = 0; i < n; ++i)
            scanf("%d%d", &machine[i].x, &machine[i].y);
        for(int i = 0; i < t; ++i)
            scanf("%d%d", &task[i].x, &task[i].y);
        sort(machine, machine + n), sort(task, task + t);
        int complete = 0;
        LL sum = 0;
        for(int i = 0, j = 0; i < t; ++i) {
            while(j < n && machine[j].x >= task[i].x) {
                tmp[machine[j].y]++;
                j++;
            }
            for (int k = task[i].y; k <= 100; ++k) {
                if(tmp[k]) {
                    ++complete;
                    sum += task[i].x * 500 * 1LL + 2 * 1LL * task[i].y;
                    tmp[k]--;
                    break;
                }
            }
        }
        printf("%d %I64d
", complete, sum);
    }
    return 0;
}

 补：

多者以上的贪心都要跟偏序挂钩