poj 1655 Balancing Act

类型：DFS

来源：POJ Monthly--2004.05.15 IOI 2003 sample task

题目：一棵树，定义每个节点的balance值：去掉这点节点后的森林里所有树的最大节点数。求出最小的balance值和其所对应的节点编号

思路：通过邻接表构造以1为根的二叉树，对其进行dfs，对于待删除节点u，其balance值为各个以u为根的子树的节点数的最大值或者除该树（u为根）以外的节点的数量。统

计每个根的balance值比较得结果

// poj 1655 Balancing Act
// 664K	47MS
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

#define FOR(i,a,b) for(i = (a); i < (b); ++i)
#define FORE(i,a,b) for(i = (a); i <= (b); ++i)
#define CLR(a,b) memset(a,b,sizeof(a))

const int MAXN = 20010;
const int INF = 0x7f7f7f7f;

int cas, n;
int cnt, minnum, endi;
bool vis[MAXN];
int head[MAXN], num[MAXN];
struct node {
int v, nxt;
}p[MAXN * 2];

void addedge(int u, int v) {
p[cnt].v = v;
p[cnt].nxt = head[u];
head[u] = cnt++;
p[cnt].v = u;
p[cnt].nxt = head[v];
head[v] = cnt++;
}

void dfs(int u) {
int i;
int maxnum = -1;

//!!!
num[u] = 1;
for(i = head[u]; i; i = p[i].nxt) {
int v = p[i].v;
if(!vis[v]) {
vis[v] = true;
dfs(v);
//!!!
num[u] += num[v];
maxnum = max(maxnum, num[v]);
}
}
maxnum = max(maxnum, n - num[u]);
if(maxnum < minnum)
minnum = maxnum, endi = u;
return;
}

int main()
{
int i, u, v;

scanf("%d", &cas);
while(cas--) {
CLR(head, 0);
CLR(vis, false);
CLR(num, 0);
cnt = 1;
minnum = INF;
scanf("%d", &n);
FOR(i, 0, n - 1) {
scanf("%d %d", &u, &v);
addedge(u, v);
}
vis[1] = true;
dfs(1);
printf("%d %d\n", endi, minnum);
}
return 0;
}