Hdu 4869 Task (Greedy) and hdu4869task greedy

Hdu 4869 Task (Greedy) and hdu4869task greedy

Link: hdu 4869 Task

There are n machines and m tasks, and each machine and task has xi and yi. The xi and yi of the machine must be greater than or equal to the xi and yi of the task to execute the task. Each machine can only execute one task per day. The number of tasks required to be completed should be as many as possible, and the amount should be as large as possible. The amount to complete each task is xi Jinping 500 + yi wei2

Solution: greedy. mach [I] [j] indicates the number of machines whose level is I and whose time is j, and task [I] [j] indicates that the level is I, the number of machines whose time is j. Each time I is preferentially reduced, because the corresponding level is reduced by 100, the corresponding amount cost will not be reduced by more than 500 (that is, the time is reduced by 1 ).
Add mach [I] [j] to mach [I] [j + 1] each time, that is, the machine not used above, tmp records the current j, the number of machines with a level greater than I.

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef __int64 ll;const int maxt = 1440;const int maxd = 100;int N, M;int mach[maxd+10][maxt+10], task[maxd+10][maxt+10];void init () {    int a, b;    memset(mach, 0, sizeof(mach));    memset(task, 0, sizeof(task));    for (int i = 0; i < N; i++) {        scanf("%d%d", &a, &b);        mach[b][a]++;    }    /*    for (int i = maxd; i >= 0; i--)        for (int j = maxt; j >= 0; j--)            mach[i][j] = mach[i][j] + mach[i+1][j] + mach[i][j+1] - mach[i+1][j+1];            */    for (int i = 0; i < M; i++) {        scanf("%d%d", &a, &b);        task[b][a]++;    }}void solve () {    ll ans = 0;    int cnt = 0;    for (int j = maxt; j >= 0; j--) {        int tmp = 0;        for (int i = maxd; i >= 0; i--) {            mach[i][j] += mach[i][j+1];            tmp += mach[i][j];            int k = min(tmp, task[i][j]);            ans += (ll)k * (2LL * i + 500LL * j);            tmp -= k;            cnt += k;            for (int x = i; x <= maxd; x++) {                int p = min(mach[x][j], k);                k -= p;                mach[x][j] -= p;                if (k == 0)                    break;            }        }    }    printf("%d %I64d\n", cnt, ans);}int main () {    while (scanf("%d%d", &N, &M) == 2 && N + M) {        init();        solve();    }    return 0;}