1650. Billionaires(线段树）

1650

简单题 线段树的单点更新 就是字符串神马的 有点小繁琐 开两个map 一个存城市 一个存名字

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cstdlib>
 #include<string>
 #include<map>
 using namespace std;
 #define N 60010
 #define LL long long
 LL s[N<<];
 LL a[N],pp[N];
 map<string,int>q;
 map<string,int>qq;
 char pq[N][];
 struct node
 {
     int d;
     char s1[];
     char s2[];
 }bi[N];
 struct mode
 {
     char ss[];
     LL vv;
 }ct[N];
 int sd[N],g,gg,lo[N];
 void up(int w)
 {
     s[w] = max(s[w<<],s[w<<|]);
 }
 void build(int l,int r,int w)
 {
     if(l==r)
     {
         s[w] = a[l];
         return ;
     }
     int m = (l+r)>>;
     build(l,m,w<<);
     build(m+,r,w<<|);
     up(w);
 }
 void update(int p,LL d,int l,int r,int w)
 {
     if(l==r)
     {
         s[w]+=d;
         return ;
     }
     int m = (l+r)>>;
     if(p<=m)
     update(p,d,l,m,w<<);
     else
     update(p,d,m+,r,w<<|);
     up(w);
 }
 int query(int l,int r,int w)
 {
     if(l==r)
     {
         return l;
     }
     int m = (l+r)>>;
     if(s[w<<]>s[w<<|])
     return query(l,m,w<<);
     else if(s[w<<]<s[w<<|])
     return query(m+,r,w<<|);
     else
     return ;
 }
 bool cmp(mode a,mode b)
 {
     return strcmp(a.ss,b.ss)<;
 }
 int main()
 {
     int n,k,m,i;
     char s1[],s2[];
     LL p;
     scanf("%d",&n);
     for(i = ; i <= n ; i++)
     {
         scanf("%s%s %lld",s1,s2,&p);
         if(!q[s2])
         {
             g++;
             q[s2] = g;
             a[q[s2]]+=p;
             strcpy(pq[g],s2);
         }
         else
         a[q[s2]]+=p;
         if(!qq[s1])
         {
             gg++;
             qq[s1] = gg;
             pp[gg] = p;
         }
         lo[qq[s1]] = q[s2];
     }
     scanf("%d%d",&m,&k);
     for(i = ; i <= k ; i++)
     {
         scanf("%d%s%s",&bi[i].d,bi[i].s1,bi[i].s2);
         if(!q[bi[i].s2])
         {
             g++;
             q[bi[i].s2] = g;
             strcpy(pq[g],bi[i].s2);
         }
     }
     build(,g,);
     int o = query(,g,);
     int t = ;
     for(i =  ; i <= k ; i++)
     {
         int o1 = qq[bi[i].s1];
         int o2 = q[bi[i].s2];
         int kk = bi[i].d;
         if(o)
         {
             sd[o]+=(kk-t);
         }
         t = kk;
         update(lo[o1],-pp[o1],,g,);
         update(o2,pp[o1],,g,);
         lo[o1] = o2;
         o = query(,g,);
     }
     if(o)
     sd[o] += m-t;
     o = ;
     for(i = ; i <= g ; i++)
     {
         if(sd[i])
         {
             strcpy(ct[o].ss,pq[i]);
             ct[o].vv = sd[i];
             o++;
         }
     }
     sort(ct,ct+o,cmp);
     for(i =  ; i < o ; i++)
     printf("%s %lld\n",ct[i].ss,ct[i].vv);
     return ;
 }