借教室 Vijos 1782 NOIP2012 D2T2 Lazy 线段树

怎么说呢，最后一个点跑了1234ms但是vijos没给TLE，我就厚颜无耻地认为自己过了吧！

标准的lazy线段树写法，权当存个模版了！

#	状态	耗时	内存占用
#1	Accepted	4ms	380.0 KiB
#2	Accepted	4ms	348.0 KiB
#3	Accepted	3ms	384.0 KiB
#4	Accepted	4ms	372.0 KiB
#5	Accepted	4ms	360.0 KiB
#6	Accepted	4ms	348.0 KiB
#7	Accepted	42ms	3.219 MiB
#8	Accepted	57ms	3.734 MiB
#9	Accepted	76ms	3.836 MiB
#10	Accepted	96ms	2.375 MiB
#11	Accepted	104ms	2.859 MiB
#12	Accepted	116ms	2.492 MiB
#13	Accepted	150ms	4.25 MiB
#14	Accepted	116ms	3.082 MiB
#15	Accepted	440ms	18.121 MiB
#16	Accepted	619ms	17.84 MiB
#17	Accepted	859ms	17.824 MiB
#18	Accepted	943ms	16.742 MiB
#19	Accepted	1185ms	18.117 MiB
#20	Accepted	1234ms	18.461 MiB

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<iostream>
 6 using namespace std;
 7 template<class T> inline void read(T &_a){
 8     bool f=0;int _ch=getchar();_a=0;
 9     while(_ch<'0' || _ch>'9'){if(_ch=='-')f=1;_ch=getchar();}
10     while(_ch>='0' && _ch<='9'){_a=(_a<<1)+(_a<<3)+_ch-'0';_ch=getchar();}
11     if(f)_a=-_a;
12 }
13 
14 const int maxn=1000001;
15 struct fff
16 {
17     int minn,lazy;
18 }node[(maxn<<2)+(maxn<<1)];
19 int n,m;
20 
21 inline void pushup(int u)
22 {
23     node[u].minn=min(node[u<<1].minn,node[u<<1|1].minn);
24 }
25 
26 inline void pushdown(int u)
27 {
28     if(node[u].lazy)
29     {
30         node[u<<1].lazy+=node[u].lazy;
31         node[u<<1|1].lazy+=node[u].lazy;
32         node[u<<1].minn-=node[u].lazy;
33         node[u<<1|1].minn-=node[u].lazy;
34         node[u].lazy=0;
35     }
36 }
37 
38 void build(int u,int l,int r)
39 {
40     if(l==r) { read(node[u].minn); return ; }
41     int mid=(l+r)>>1;
42     build(u<<1,l,mid);
43     build(u<<1|1,mid+1,r);
44     pushup(u);
45 }
46 
47 int query(int u,int l,int r,int L,int R)
48 {
49     if(L<=l&&r<=R) return node[u].minn;
50     pushdown(u);
51     int mid=(l+r)>>1;
52     if(R<=mid) return query(u<<1,l,mid,L,R);
53     if(L>mid) return query(u<<1|1,mid+1,r,L,R);
54     return min(query(u<<1,l,mid,L,R),query(u<<1|1,mid+1,r,L,R));
55 }
56 
57 void update(int u,int l,int r,int L,int R,int val)
58 {
59     if(L<=l&&r<=R)
60     {
61         node[u].lazy+=val;
62         node[u].minn-=val;
63         return ;
64     }
65     pushdown(u);
66     int mid=(l+r)>>1;
67     if(L<=mid) update(u<<1,l,mid,L,R,val);
68     if(R>mid) update(u<<1|1,mid+1,r,L,R,val);
69     pushup(u);
70     return ;
71 }
72 
73 inline void out()
74 {
75     for (register int i=1;i<=n;++i)
76     printf("%d ",query(1,1,n,i,i));putchar('
');
77 }
78 
79 int main()
80 {
81     read(n); read(m);
82     build(1,1,n);
83     // out();
84     for (register int cas=1,d,s,j;cas<=m;++cas)
85     {
86         read(d); read(s); read(j);
87         int num=query(1,1,n,s,j);
88         // printf("[%d]
",num);
89         if(num<d) {printf("-1
%d",cas); return 0;}
90         update(1,1,n,s,j,d);
91         // out();
92     }
93     printf("0");
94     return 0;
95 }

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原文地址：https://www.cnblogs.com/jaywang/p/7726375.html