poj3140 Contestants Division(树形DP)

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思路：f[u] 为以u为结点，它的子孙的所有结点数,因为sum为树的全部结点数，所以 sum - f[u] 为除了u结点这棵树剩下的结点数。这样的话求差，就求得了 u结点和他的父节点这条边断开后两个连通图的差值。

注意: 要用long long, 最大值要取到10^9 * n; abs自己写。

附f[i] 和 sum - f[i]的关系图一张，方便理解。

 /*********************************************** * Author: fisty * Created Time: 2015/2/22 20:53:05 * File Name   : poj3140.cpp *********************************************** */#include <iostream>#include <cstring>#include <deque>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <bitset>#include <algorithm>using namespace std;#define Debug(x) cout << #x << " " << x <<endl#define Memset(x, a) memset(x, a, sizeof(x))const int INF = 0x3f3f3f3f;typedef long long LL;typedef pair<int, int> P;#define FOR(i, a, b) for(int i = a;i < b; i++)#define MAX_N 100100#define ABS(x) x < 0 ? (-(x)) : x#define MIN(a, b) a < b ? a : bint n, m;LL sum;LL f[MAX_N];LL ans;vector<int> G[MAX_N];void dfs(int u, int fa){    for(int i = 0;i < G[u].size(); i++){        int v = G[u][i];        if(v == fa) continue;        dfs(v, u);        f[u] += f[v];    }    LL _min = ABS(sum - f[u] - f[u]);    //Debug(_min);    ans = MIN(ans, _min);}int main() {    //freopen("in.cpp", "r", stdin);    int cnt = 1;    while(scanf("%d%d", &n, &m)){        Memset(f, 0);        if(!n && !m) break;        sum = 0;        FOR(i, 1, n+1){            G[i].clear();            scanf("%I64d", &f[i]);            sum += f[i];        }        FOR(i, 0, m){            int u, v;            scanf("%d%d", &u, &v);            G[u].push_back(v);            G[v].push_back(u);        }        ans = 11000000000000;        dfs(1, -1);        printf("Case %d: %I64d\n", cnt++, ans);    }    return 0;}

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