离线树状数组 hihocoder 1391 Countries

Posted 2020-08-11 blues

官方题解：

 

 1 // 离线树状数组 hihocoder 1391 Countries
 2 
 3 #include <iostream>
 4 #include <cstdio>
 5 #include <cstdlib>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <math.h>
 9 #include <memory.h>
10 using namespace std;
11 #define LL long long
12 typedef pair<int,int> pii;
13 const LL inf = 0x3f3f3f3f;
14 const LL MOD =100000000LL;
15 // const int N = 1e5+10;
16 const double eps = 1e-8;
17 void fre() {freopen("in.txt","r",stdin);}
18 void freout() {freopen("out.txt","w",stdout);}
19 inline int read() {int x=0,f=1;char ch=getchar();while(ch>\'9\'||ch<\'0\') {if(ch==\'-\') f=-1; ch=getchar();}while(ch>=\'0\'&&ch<=\'9\') {x=x*10+ch-\'0\';ch=getchar();}return x*f;}
20 const int MAXN = 60010;
21 struct node{
22     LL l,r,v;
23 }N[MAXN];
24 
25 LL f[MAXN],x[MAXN];
26 int cmp(node a,node b){
27     return a.r<b.r;
28 }
29 void add(LL w,LL v){
30     for(;w<=MAXN;w+=w&(-w)) f[w]+=v;
31 }
32 LL getsum(LL w){
33     LL sum=0;
34     for(;w;w-=w&(-w)) sum+=f[w];
35     return sum;
36 }
37 
38 int main(){
39     LL n,m,bs,nn,M;
40     while(~scanf("%lld%lld",&n,&m)){
41         memset(f,0,sizeof(f));
42         memset(x,0,sizeof(x));
43         scanf("%lld",&bs);
44         scanf("%lld%lld",&nn,&M);
45         LL r=m+bs;
46         LL k=0;
47         LL sum=0;
48         for(LL i=1;i<=nn;i++){
49             LL s,t,v;
50             scanf("%lld%lld%lld",&s,&t,&v);
51             if(s+t<bs||s+t>r) continue;
52             else{
53                 N[++k].l=s+2*t,N[k].r=s+2*t+(r-s-t)/(2*t)*2*t;
54                 N[k].v=v;
55                 sum+=v;
56             }
57         }
58         for(LL i=1;i<=M;i++){
59             LL s,t,v;
60             scanf("%lld%lld%lld",&s,&t,&v);
61             N[++k].l=s+t,N[k].v=v;
62             sum+=v;
63             if(s+t*2<bs||s+t*2>r) N[k].r=N[k].l;
64             else{
65                 N[k].r=s+3*t+(r-s-2*t)/(2*t)*2*t;
66             }
67         }
68         LL g=k;
69         k=0;
70         for(LL i=1;i<=g;i++){
71             x[++k]=N[i].l,x[++k]=N[i].r,x[++k]=N[i].r-n;
72         }
73         sort(N+1,N+1+g,cmp);
74         sort(x+1,x+1+k);
75         k=unique(x+1,x+1+k)-x-1;
76         LL ans=-inf;
77         for(LL i=1;i<=g;i++){
78             LL w,pre,now;
79             w=lower_bound(x+1,x+1+k,N[i].l)-x;
80             now=lower_bound(x+1,x+1+k,N[i].r)-x;
81             pre=lower_bound(x+1,x+1+k,N[i].r-n)-x;
82             add(w,N[i].v);
83             ans=max(ans,getsum(now)-getsum(pre-1));
84         }
85         printf("%lld\\n",sum-ans);
86     }
87     return 0;
88 }