P1083 借教室

Posted 2020-10-09 Herminone

经典题

二分 + 前缀和

注意long long 和初始化change[]      //QAQ

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #define ll long long   //注意long long
 6 
 7 using namespace std;
 8 const int N = 1000000 + 5;
 9 ll n,m;
10 ll d[N],sum[N],c[N],a[N],b[N],change[N];
11 bool bo;

12 bool check(ll x){
13     memset(change,0,sizeof(change));     //注意初始化QAQ
14     for(int i = 1;i <= x;++ i){
15         change[a[i]] += c[i];
16         change[b[i] + 1] -= c[i];
17     }
18     ll sum = 0;
19     for(int i = 1;i <= n;++ i){
20         sum += change[i];
21         if(sum > d[i]) return true;
22     }
23     return false;
24 }
25 int main(){
26     scanf("%lld%lld",&n,&m);
27     for(int i = 1;i <= n;++ i){
28         scanf("%lld",&d[i]);
29     }
30     for(int i = 1;i <= m;++ i){
31         scanf("%lld%lld%lld",&c[i],&a[i],&b[i]);
32     }
33     ll l = 0,r = m;  // l :不可能，r:可能 
34     while(r - l > 1){
35         ll mid = (l + r) >> 1;
36         if(check(mid)){
37             r = mid;
38             bo = 1;  //有输出的可能值
39         }
40         else{
41             l = mid;
42         }
43     }
44     if(bo){
45         cout << "-1" << ‘\n‘;
46         cout << r << ‘\n‘;  //输出可能值r
47     }
48     else{
49         cout << "0" << ‘\n‘;
50     }
51     return 0;
52 }