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

  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<cstdlib>
  6 #include<string>
  7 #include<map>
  8 using namespace std;
  9 #define N 60010
 10 #define LL long long
 11 LL s[N<<2];
 12 LL a[N],pp[N];
 13 map<string,int>q;
 14 map<string,int>qq;
 15 char pq[N][25];
 16 struct node
 17 {
 18     int d;
 19     char s1[25];
 20     char s2[25];
 21 }bi[N];
 22 struct mode
 23 {
 24     char ss[25];
 25     LL vv;
 26 }ct[N];
 27 int sd[N],g,gg,lo[N];
 28 void up(int w)
 29 {
 30     s[w] = max(s[w<<1],s[w<<1|1]);
 31 }
 32 void build(int l,int r,int w)
 33 {
 34     if(l==r)
 35     {
 36         s[w] = a[l];
 37         return ;
 38     }
 39     int m = (l+r)>>1;
 40     build(l,m,w<<1);
 41     build(m+1,r,w<<1|1);
 42     up(w);
 43 }
 44 void update(int p,LL d,int l,int r,int w)
 45 {
 46     if(l==r)
 47     {
 48         s[w]+=d;
 49         return ;
 50     }
 51     int m = (l+r)>>1;
 52     if(p<=m)
 53     update(p,d,l,m,w<<1);
 54     else
 55     update(p,d,m+1,r,w<<1|1);
 56     up(w);
 57 }
 58 int query(int l,int r,int w)
 59 {
 60     if(l==r)
 61     {
 62         return l;
 63     }
 64     int m = (l+r)>>1;
 65     if(s[w<<1]>s[w<<1|1])
 66     return query(l,m,w<<1);
 67     else if(s[w<<1]<s[w<<1|1])
 68     return query(m+1,r,w<<1|1);
 69     else
 70     return 0;
 71 }
 72 bool cmp(mode a,mode b)
 73 {
 74     return strcmp(a.ss,b.ss)<0;
 75 }
 76 int main()
 77 {
 78     int n,k,m,i;
 79     char s1[25],s2[25];
 80     LL p;
 81     scanf("%d",&n);
 82     for(i = 1; i <= n ; i++)
 83     {
 84         scanf("%s%s %lld",s1,s2,&p);
 85         if(!q[s2])
 86         {
 87             g++;
 88             q[s2] = g;
 89             a[q[s2]]+=p;
 90             strcpy(pq[g],s2);
 91         }
 92         else
 93         a[q[s2]]+=p;
 94         if(!qq[s1])
 95         {
 96             gg++;
 97             qq[s1] = gg;
 98             pp[gg] = p;
 99         }
100         lo[qq[s1]] = q[s2];
101     }
102     scanf("%d%d",&m,&k);
103     for(i = 1; i <= k ; i++)
104     {
105         scanf("%d%s%s",&bi[i].d,bi[i].s1,bi[i].s2);
106         if(!q[bi[i].s2])
107         {
108             g++;
109             q[bi[i].s2] = g;
110             strcpy(pq[g],bi[i].s2);
111         }
112     }
113     build(1,g,1);
114     int o = query(1,g,1);
115     int t = 0;
116     for(i = 1 ; i <= k ; i++)
117     {
118         int o1 = qq[bi[i].s1];
119         int o2 = q[bi[i].s2];
120         int kk = bi[i].d;
121         if(o)
122         {
123             sd[o]+=(kk-t);
124         }
125         t = kk;
126         update(lo[o1],-pp[o1],1,g,1);
127         update(o2,pp[o1],1,g,1);
128         lo[o1] = o2;
129         o = query(1,g,1);
130     }
131     if(o)
132     sd[o] += m-t;
133     o = 0;
134     for(i = 1; i <= g ; i++)
135     {
136         if(sd[i])
137         {
138             strcpy(ct[o].ss,pq[i]);
139             ct[o].vv = sd[i];
140             o++;
141         }
142     }
143     sort(ct,ct+o,cmp);
144     for(i = 0 ; i < o ; i++)
145     printf("%s %lld\n",ct[i].ss,ct[i].vv);
146     return 0;
147 }
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posted @ 2013-10-01 21:36  _雨  阅读(428)  评论(0)  编辑  收藏  举报