离线树状数组 hihocoder 1391 Countries

官方题解：

 // 离线树状数组 hihocoder 1391 Countries

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <algorithm>
 #include <vector>
 #include <math.h>
 #include <memory.h>
 using namespace std;
 #define LL long long
 typedef pair<int,int> pii;
 const LL inf = 0x3f3f3f3f;
 const LL MOD =100000000LL;
 // const int N = 1e5+10;
 const double eps = 1e-;
 void fre() {freopen("in.txt","r",stdin);}
 void freout() {freopen("out.txt","w",stdout);}
 inline int read() {int x=,f=;char ch=getchar();while(ch>''||ch<'') {if(ch=='-') f=-; ch=getchar();}while(ch>=''&&ch<='') {x=x*+ch-'';ch=getchar();}return x*f;}
 const int MAXN = ;
 struct node{
     LL l,r,v;
 }N[MAXN];

 LL f[MAXN],x[MAXN];
 int cmp(node a,node b){
     return a.r<b.r;
 }
 void add(LL w,LL v){
     for(;w<=MAXN;w+=w&(-w)) f[w]+=v;
 }
 LL getsum(LL w){
     LL sum=;
     for(;w;w-=w&(-w)) sum+=f[w];
     return sum;
 }

 int main(){
     LL n,m,bs,nn,M;
     while(~scanf("%lld%lld",&n,&m)){
         memset(f,,sizeof(f));
         memset(x,,sizeof(x));
         scanf("%lld",&bs);
         scanf("%lld%lld",&nn,&M);
         LL r=m+bs;
         LL k=;
         LL sum=;
         for(LL i=;i<=nn;i++){
             LL s,t,v;
             scanf("%lld%lld%lld",&s,&t,&v);
             if(s+t<bs||s+t>r) continue;
             else{
                 N[++k].l=s+*t,N[k].r=s+*t+(r-s-t)/(*t)**t;
                 N[k].v=v;
                 sum+=v;
             }
         }
         for(LL i=;i<=M;i++){
             LL s,t,v;
             scanf("%lld%lld%lld",&s,&t,&v);
             N[++k].l=s+t,N[k].v=v;
             sum+=v;
             if(s+t*<bs||s+t*>r) N[k].r=N[k].l;
             else{
                 N[k].r=s+*t+(r-s-*t)/(*t)**t;
             }
         }
         LL g=k;
         k=;
         for(LL i=;i<=g;i++){
             x[++k]=N[i].l,x[++k]=N[i].r,x[++k]=N[i].r-n;
         }
         sort(N+,N++g,cmp);
         sort(x+,x++k);
         k=unique(x+,x++k)-x-;
         LL ans=-inf;
         for(LL i=;i<=g;i++){
             LL w,pre,now;
             w=lower_bound(x+,x++k,N[i].l)-x;
             now=lower_bound(x+,x++k,N[i].r)-x;
             pre=lower_bound(x+,x++k,N[i].r-n)-x;
             add(w,N[i].v);
             ans=max(ans,getsum(now)-getsum(pre-));
         }
         printf("%lld\n",sum-ans);
     }
     return ;
 }