Codeforces 238C World Eater

Test instructions

Enter N, indicating that there are N cities, respectively, 1~n, enter the N-1 group, A, a, indicates that from a can go to B, the minimum number of changes to change the road from the two of the city can go through all the city.

(Http://blog.csdn.net/metalseed/article/details/8045038%20 of something related to achievement)

Problem Solving Ideas:

Create a tree, break an edge, divide the tree into two parts, and calculate the minimum amount of change in both parts.

Then it becomes the smallest change in the subtree.

Dp[i] Indicates the amount of change required for a subtree with I as root

If Root is the starting point, then cost[root] = Dp[root] If you start with any of these subtrees trees

Cost[u] = Dp[root] + u-root need to change amount-root-u need to change the amount of

To make Cost[u] minimum, then (U-root need to change +root-u need to change) the minimum value

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5#include <cmath>6 using namespacestd;7 Const intN = 1e5 +Ten;//1e5 is 10 of the 5-time Square8 structEdge9 {Ten     int  from, to, DIS, NEX; One}edge[n<<1]; A intHead[n], edgenum; - voidAddintUintVintd) - { theEdge E ={u, V, D, Head[u]}; -Edge[edgenum] =E; -Head[u] = edgenum++; - } + intDp[n]; - intn, MX; + voidDfsintUintFaintval) A { atDp[u] =0; -      for(inti = Head[u]; ~i; i = Edge[i].nex)//~i said to take the reverse, in fact, is not 0 -     { -         intv =edge[i].to; -         if(v = = FA)Continue; -DFS (V, u, val-Edge[i].dis);  inDp[u] + = Dp[v] + (edge[i].dis!=1);//if the parentheses are set, add 1, otherwise add 0 . -     } toMX =Max (MX, val); + } - intMain () the { *      while(Cin >>N) $     {Panax Notoginseng         if(n = =1) {puts ("0");Continue; }//puts output and change the costume file Cstdio -Memset (Head,-1,sizeofhead); theEdgenum =0; +          for(inti =1, u, v; I < n; i++) A         { thescanf"%d%d", &u, &v); +Add (U, V,1); -Add (V, U,-1);  $         } $         intAns =1<< -; -          for(inti =0, u, v; i < Edgenum; i + =2) -         { theU = edge[i]. from; v =edge[i].to; -MX =-(1<< -);WuyiDFS (U, V,0); the             intTMP = Dp[u]-MX; -MX =-(1<< -); WuDFS (V, u,0); -TMP + = Dp[v]-MX; AboutAns =min (ans, tmp); $         } -printf"%d\n", ans); -     } -     return 0; A}

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Codeforces 238C World Eater