树的平衡值

POJ 1655

题目描述：

树的顶点~20,000. 保证是一棵树,每个顶点的平衡值如下:去掉该点后,剩下的树顶点的最大值.求出给出树平衡值最小的顶点,如果一样,输出序号较小的.

题解：

树形dp.其实平衡值就是儿子的个数和父亲那里个数的较小值.

重点：

树形dp,统计子树个数经常用.

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <ctype.h>#include <limits.h>#include <cstdlib>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#include <set>#include <bitset>#define CLR(a) memset(a, 0, sizeof(a))#define REP(i, a, b) for(int i = a;i < b;i++)#define REP_D(i, a, b) for(int i = a;i <= b;i++)typedef long long ll;using namespace std;const int maxn = 2e4 +100;const int INF = INT_MAX - 1;int num[maxn], dp[maxn], n;int ans, ans_i;vector<int> G[maxn];void dfs(int u, int fa){    num[u] = 1;    dp[u] = 0;    REP(i, 0, G[u].size())//一边搞结果dp,一遍算num,num才是真正的树形dp数组.    {        int v = G[u][i];        if(v!=fa)        {            dfs(v, u);            dp[u] = max(dp[u], num[v]);            num[u] += num[v];        }    }    dp[u] = max(dp[u], n - num[u]);    if(dp[u] < ans)    {        ans = dp[u];        ans_i = u;    }    else if(dp[u] == ans && u < ans_i)    {        ans_i = u;    }}void solve(){    ans = INF;    dfs(1, 0);    printf("%d %d\n", ans_i, ans);}int main(){    //freopen("1Ain.txt", "r", stdin);    //freopen("1Aout.txt", "w", stdout);    int ncase;    scanf("%d", &ncase);    while(ncase--)    {        scanf("%d", &n);        REP_D(i, 1, n)        {            G[i].clear();        }        REP_D(i, 1, n - 1)        {            int a, b;            scanf("%d%d", &a, &b);            G[a].push_back(b);//            if(a == 1)//            {//                printf("%d--%d\n",i, b);//            }            G[b].push_back(a);//            if(b == 1)//            {//                printf("%d--%d\n", i,a);//            }        }        solve();    }    return 0;}