Cattle-off practice match 56 B Chen and his small school

Topic links: https://ac.nowcoder.com/acm/contest/3566/B

Ideas: a road can be divided into the left and right portions of FIG.

l_ci, l_peo, r_ci, r_peo, w respectively, left the city and the number of persons, the right number and the number of cities, the cost of the road.

We know that people go to the left r_ci cities on the right, the right of people to go l_ci cities left,

Then the cost of the path is cost = 2 * w * (l_ci * r_peo + r_ci * l_peo); (note that back and forth).

We can use the topological sort to derive an edge around the city and the number of cases.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <queue>
 5 using namespace std;
 6 typedef long long ll;
 7 
 8 const int N = (int)2e5+100;
 9 struct node{
10     int to,nxt;
11 }e[N<<1];
12 struct Info{
13     int ci,peo;
14     Info(){ci = peo = 0;}
 15  } info [N];
 16  int A [N], W is [N], du [N], head [N]; // number, toll, degrees 
. 17  BOOL VIS [N];
 18 is  int n-, U , V, W, TOT; LL sum_peo;
 . 19  
20 is inline void the Add ( int U, int V) {
 21 is      E [TOT] .to = V; E [TOT] .nxt head = [U]; head [U] = ++ TOT ;
 22 is      E [TOT] .to = U; E [TOT] = .nxt head [V]; head [V] TOT = ++ ;
 23 is  }
 24  
25  void top_sort () {
 26 is      // degree of pressure into a 
27      Queue < int >que;
 28      for ( int I = . 1 ; I <= n-; ++ I) {
 29          IF (du [I] == . 1 ) {
 30              VIS [I] = . 1 ;
 31 is              que.push (I);
 32          }
 33 is      }
 34 is  
35      the while (! {que.empty ())
 36          int now = que.front (); que.pop ();
 37 [          // add their city case 
38 is          info [now] + = .ci . 1 ;
 39          info [now] + = .peo A [now];
 40          for( Int O = head [now]; ~ O; O = E [O] .nxt) {
 41 is              int to = E [O] .to;
 42 is              IF (! {VIS [to])
 43 is                  // pass their city the case 
44 is                  info [to] + = .ci info [now] .ci;
 45                  info [to] + = .peo info [now] .peo;
 46 is                  IF (--du [to] == . 1 ) {
 47                      VIS [ to] = . 1 ;
 48                      que.push (to);
 49                  }
 50              }
 51 is          }
 52 is      }
53 }
54 
55 void show_info(){
56     for(int i = 1;i <= n; ++i){
57         printf("now = %d city = %d people = %d\n",i,info[i].ci,info[i].peo);
58     }
59 }
60 
61 inline ll cost(Info& u,int i){
62     Info v;
63     v.ci = n-u.ci;
64     v.peo = sum_peo - u.peo;
65   //  printf("u.c = %d v.c = %d  u.p = %d v.p = %d W = %d\n",u.ci,v.ci,u.peo,v.peo,W[i]);
66     return (ll)2*W[i]*((ll)u.ci*v.peo+(ll)v.ci*u.peo);
67 }
68 
69 int main(){
70 
71     scanf("%d",&n);
72     for(int i = 0; i <= n; ++i) head[i] = -1; tot = 0;
73     for(int i = 1; i <= n; ++i){
74         scanf("%d",A+i);
75         sum_peo += A[i];
76     }
77     for(int i = 1; i < n; ++i){
78         scanf("%d%d%d",&u,&v,W+i);
79         add(u,v);
80         ++du[u]; ++du[v];//无向边
81     }
82     top_sort();
83     //show_info();
84     for(int i = 0; i < 2*(n-1); i+=2 ) {
 85          // we always selected from a small number of cities edge drawn aside and the other side of the city where the number 
86          int X = info [E [I] .to] .ci <info [E [I ^ . 1 ?] .to] .ci E [I] .to: E [I ^ . 1 ] .to;
 87       //    the printf ( "City D =% \ n-", X); 
88          the printf ( " % LLD \ n- " , cost (info [X], . 1 + (I >> . 1 )));
 89      }
 90  
91 is      return  0 ;
 92 }