POJ 3107 Godfather (tree-shaped DP)

Title Address: POJ 3107

Tree-shaped DP water problem. Records the maximum number of nodes for each sub-tree of each point, as well as the summary points for all subtrees. And then traverse through to find the line.

The code is as follows:

#include <iostream> #include <string.h> #include <math.h> #include <queue> #include < algorithm> #include <stdlib.h> #include <map> #include <set> #include <stdio.h>using namespace std; #define LL __int64#define Pi ACOs ( -1.0) const int Mod=100000000;const int Inf=0x3f3f3f3f;const double eqs=1e -8;int dp[60000], head[60000], CNT, tot[60000], c[60000], num, min1;struct node {int u, V, next;} edge[120000];voi        d Add (int u, int v) {edge[cnt].v=v;        Edge[cnt].next=head[u]; head[u]=cnt++;}        void Dfs (int u, int fa) {tot[u]=1;                for (int i=head[u]; i!=-1; i=edge[i].next) {int v=edge[i].v;                if (V==FA) continue;                DFS (V,U);                Dp[u]=max (Dp[u],tot[v]);        TOT[U]+=TOT[V];        }}void init () {memset (head,-1,sizeof (head));        Memset (Dp,0,sizeof (DP));        cnt=0;        num=0; Min1=inf;}        int main () {int n, I, J, u, V; while (sCANF ("%d", &n)!=eof) {init ();                        for (I=1; i<n; i++) {scanf ("%d%d", &u,&v);                        Add (U,V);                Add (V,u);                } dfs (1,-1);                        for (I=1; i<=n; i++) {Dp[i]=max (dp[i],n-tot[i]);                        if (Min1>dp[i]) {min1=dp[i];                                }} for (I=1; i<=n; i++) {if (dp[i]==min1) {                        C[num++]=i;                        }} for (i=0; i<num; i++) {printf ("%d", c[i]);                if (i!=num-1) printf ("");        } puts (""); } return 0;}

POJ 3107 Godfather (tree-shaped DP)