HDU 4864 Task (Greedy)

HDU 4864 Task (Greedy)
HDU 4864 Task

Question Link

Some machines and tasks all have time and level. The condition for a machine to perform a task is that the time level is greater than or equal to the task, and a task can only be performed by one machine, now we want to maximize the task and ensure as much money as possible.

Idea: greedy. For each task, priority is given to matching when the time is large. If the time is the same, priority is given to matching when the level is large. Because time accounts for a large amount of time, when the time is 1, it will be 500 more, the maximum level is 200. When matching, try to use a match with a lower level, and the time can be used as long as it is greater than it, because it is time-based, so if the time can match a larger, other matches can certainly be matched, so the priority must be greater, so as long as the level is as small as possible.

Code:

#include 
 
  #include 
  
   #include 
   
    using namespace std;typedef __int64 ll;const int N = 1444;const int M = 105;int n, m;ll mac[N][M], task[N][M], used[N];int main() {    while (~scanf("%d%d", &n, &m)) {int x, y;memset(mac, 0, sizeof(mac));memset(task, 0, sizeof(task));memset(used, 0, sizeof(used));for (int i = 0; i < n; i++) {    scanf("%d%d", &x, &y);    mac[x][y]++;}for (int i = 0; i < m; i++) {    scanf("%d%d", &x, &y);    task[x][y]++;}for (int i = 0; i <= 100; i++) {    for (int j = 1439; j > 0; j--) {mac[j][i] += mac[j + 1][i];    }}ll num = 0;ll ans = 0;for (ll i = 1439; i > 0; i--) {    for (ll j = 100; j >= 0; j--) {if (!task[i][j]) continue;for (ll k = j; k <= 100; k++) {    if (task[i][j] > mac[i][k] - used[k]) {num += mac[i][k] - used[k];ans += (mac[i][k] - used[k]) * (i * 500 + j * 2);task[i][j] -= (mac[i][k] - used[k]);used[k] = mac[i][k];    }    else {num += task[i][j];ans += task[i][j] * (i * 500 + j * 2);used[k] += task[i][j];task[i][j] = 0;break;    }}    }}printf("%I64d %I64d\n", num, ans);    }    return 0;}