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  • 离线树状数组 hihocoder 1391 Countries

    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
    ",sum-ans);
    86     }
    87     return 0;
    88 }
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  • 原文地址:https://www.cnblogs.com/ITUPC/p/5911947.html
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