[NOIP2012 提高组] 借教室

题意

学校在n天内每天有ai个教室可以租借,现在有m个订单,每个订单需要在第si天至第ti天租借di个教室,现在按顺序处理订单,判断能否满足所有订单,若不行,求第几个订单开始不满足

解题思路:

1.要让区间减去某个值,可以构造差分数组来处理
2.求第几个订单开始不满足,满足二分解答适用的情况

代码如下:

#include <bits/stdc++.h>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1000005
ll a[N],b[N];
struct P {
    int d, x, y;
}p[N];
int n, m;
bool check(int mid) {
    memset(b, 0, sizeof(b));
    //差分来保存区间的减少值
    for (int j = 1; j <= mid; j++) {
        b[p[j].x] -= p[j].d;b[p[j].y+1]+=p[j].d;
    }
    for (int i = 1; i <= n; i++) {
        b[i] += b[i - 1];
        if (a[i]+b[i] < 0)return 1;
    }
    return 0;
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)cin >> a[i];
    for (int j = 1; j <= m; j++) {
        cin >> p[j].d >> p[j].x >> p[j].y;
    }
    //二分满足到第几个订单
    int l = 1, r = m+1;
    while (l < r) {
        int mid = l + r >> 1;
        if (check(mid))r = mid;
        else l = mid + 1;
    }
    if (l == m + 1)cout << 0;
    else cout << -1 << "\n" << l;
}