POJ 3140 Contestants Division (树dp)

题目链接：http://poj.org/problem?id=3140

题意：

        给你一棵树，问你删去一条边，形成的两棵子树的节点权值之差最小是多少。

思路：

        dfs

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int N = 1e5 + 5;
typedef long long LL;
int n, cnt, head[N];
LL a[N];
struct Edge {
    int next, to;
}edge[N << 1];
LL dp[N];

inline void add(int u, int v) {
    edge[cnt].next = head[u];
    edge[cnt].to = v;
    head[u] = cnt++;
}

LL dfs(int u, int p) {
    dp[u] = a[u];
    for(int i = head[u]; ~i; i = edge[i].next) {
        int v = edge[i].to;
        if(v == p)
            continue;
        dp[u] += dfs(v, u);
    }
    return dp[u];
}

LL f(LL a) {
    return a > 0 ? a : -a;
}

int main()
{
    int m, ca = 1;
    while(~scanf("%d %d", &n, &m) && (n || m)) {
        LL sum = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%lld", a + i);
            sum += a[i];
            head[i] = -1;
        }
        cnt = 0;
        int u, v;
        while(m--) {
            scanf("%d %d", &u, &v);
            add(u, v);
            add(v, u);
        }
        dfs(1, -1);
        LL res = sum;
        for(int i = 1; i <= n; ++i) {
            res = min(res, f(sum - 2*dp[i]));
        }
        printf("Case %d: %lld\n", ca++, res);
    }
    return 0;
}