luogu P1083 借教室

Posted 2021-01-18 yuyanjiab

传送门

小水题吧

二分能处理到的询问即可

用差分维护前x个订单之后的值

最后求一遍前缀和 如果爆负就是有不满足的

复杂度O((m+n)lgm)

或者区间加和区间最小值线段树也行(常数略大)

Code:(线段树)

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #include <iostream>
 5 #include<algorithm>
 6 #define rep(i,a,n) for(int i = a;i <= n;++i)
 7 #define per(i,n,a) for(int i = n;i >= a;--i)
 8 #define inf 2147483647
 9 #define ms(a,b) memset(a,b,sizeof a)
10 using namespace std;
11 typedef long long ll;
12 ll read() {
13     ll as = 0,fu = 1;
14     char c = getchar();
15     while(c < ‘0‘ || c > ‘9‘) {
16     if(c == ‘-‘) fu = -1;
17     c = getchar();
18     }
19     while(c >= ‘0‘ && c <= ‘9‘) {
20     as = as * 10 + c - ‘0‘;
21     c = getchar();
22     }
23     return as * fu;
24 }
25 //head
26 const int N = 1000006;
27 int n,m;
28 int a[N];
29 
30 #define ls t<<1
31 #define rs t<<1|1
32 int p[N<<2],add[N<<2];
33 void pdown(int l,int r,int t) {
34     if(!add[t]) return;
35     p[ls] += add[t],p[rs] += add[t];
36     add[ls] += add[t],add[rs] += add[t];
37     add[t] = 0;
38 }
39 void build(int l,int r,int t) {
40     if(l == r) return void(p[t] = read());
41     int m = l+r >> 1;
42     build(l,m,ls),build(m+1,r,rs);
43     p[t] = min(p[ls],p[rs]);
44 }
45 void update(int L,int R,int c,int l,int r,int t) {
46     if(L <= l && r <= R) {
47     p[t] -= c;
48     add[t] -= c;
49     return;
50     }
51     pdown(l,r,t);
52     int m = l+r >> 1;
53     if(L <= m) update(L,R,c,l,m,ls);
54     if(R >  m) update(L,R,c,m+1,r,rs);
55     p[t] = min(p[ls],p[rs]);
56 }
57 int query(int L,int R,int l,int r,int t) {
58     if(L <= l && r <= R) return p[t];
59     pdown(l,r,t);
60     int m = l+r >> 1;
61     int ans = inf;
62     if(L <= m) ans = min(ans,query(L,R,l,m,ls));
63     if(R >  m) ans = min(ans,query(L,R,m+1,r,rs));
64     return ans;
65 }
66 
67 int main() {
68     n = read(),m = read();
69     build(1,n,1);
70     rep(i,1,m) {
71     int d = read(),x = read(),y = read();
72     update(x,y,d,1,n,1);
73     if(query(x,y,1,n,1) < 0) {
74         printf("-1
%d
",i);
75         return 0;
76     }
77     }
78     puts("0");
79     return 0;
80 }