POJ 1655 樹的重心

題目大意

多組數據，給出一棵樹，求刪除一個節點後剩餘聯通節點的最大值最小的節點。

思路

就是求重心。很簡單的一道題，1A

代碼

#include<cstdio>
#include<cstring>
#define emax(a,b) ((a) < (b) ? (b) : (a))
using namespace std;
const int MAXN = 20005;
const int MAXE = 40005;
const int INF = 0x3f3f3f3f;
struct edge{
    int v,next;
};
edge e[MAXE];
int head[MAXN],subTree[MAXN],dp[MAXN];
int T,k,n,maxw,maxu;
char inc;

inline void adde(int u,int v){
    e[k].v = v;
    e[k].next = head[u];
    head[u] = k++;
}

inline void read(int & x){
    x = 0;
    inc = getchar();
    while(inc < '0' || inc > '9'){
        inc = getchar();
    }
    while(inc >= '0' && inc <= '9'){
        x = (x << 3) + (x << 1) + (inc ^ 48);
        inc = getchar();
    }
}

void dfs(int u,int fa){
    subTree[u] = 1;
    for(int i = head[u];~i;i = e[i].next){
        int v = e[i].v;
        if(v == fa) continue;
        dfs(v,u);
        subTree[u] += subTree[v];
        dp[u] = emax(dp[u],subTree[v]); 
    }
    dp[u] = emax(dp[u],n - subTree[u]);
}

int main(){
    read(T);
    while(T--){
        read(n);
        memset(head,-1,sizeof(head));
        k = 1;
        memset(dp,0,sizeof(dp));
        for(int i = 2;i <= n;i++){
            int u,v;
            read(u);read(v);
            adde(u,v);
            adde(v,u);
        }
        dfs(1,0);
        maxw = INF;
        maxu = -1;
        for(int i = 1;i <= n;i++){
            if(dp[i] < maxw){
                maxu = i;
                maxw = dp[i];
            }
        }
        printf("%d %d\n",maxu,maxw);
    }
    return 0;
}