POJ 3140 Contestants Division

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树形dp，水。求将一棵节点有权值的树分成两块，两块总权值差最小值。

ps：注意本题要用longlong，第一次应该wa在这里，然后换了种写法，导致tot和mv未初始化，再wa一发。还是将tot和mv的初始化放进init中好，这类题目初始化问题需要格外注意，不能忘了调用init。

#include <cstdio>#include <algorithm>#include <cstring>#include <cstdlib>#include <iostream>#include <cmath>using namespace std;typedef long long LL;#define INF 100000000000007#define N 200005int n, m;struct Node{    int v;    Node *next;}e[N], *head[N];LL sum[N], peo[N];int ptr = 0;LL mv;LL tot;void init(){    ptr = 0;    mv = INF;    tot = 0;    for(int i = 0; i <= n; i++){        head[i] = NULL;        sum[i] = 0;        peo[i] = 0;    }}void adde(int x, int y){    e[ptr].v = x;    e[ptr].next = head[y];    head[y] = &e[ptr ++];    e[ptr].v = y;    e[ptr].next = head[x];    head[x] = &e[ptr ++];}LL tdp(int x, int par){    Node *p = head[x];    sum[x] += peo[x];    while(p){        if(p -> v != par){            LL vv = tdp(p -> v, x);            sum[x] += vv;        }        p = p -> next;    }    return sum[x];}int main(){    int num = 1;    while(scanf("%d%d", &n, &m) , n || m){        init();        for(int i = 1; i <= n; i++){            scanf("%d", &peo[i]);            tot += peo[i];        }        int x, y;        for(int i = 0; i < m; i++){            scanf("%d%d", &x, &y);            adde(x, y);        }        tdp(1, 0);        for(int i = 1; i <= n; i++){            LL vv = tot - sum[i] * 2;            if(vv < 0) vv = - vv;            if(vv < mv)mv = vv;        }        printf("Case %d: ", num++);        printf("%I64d\n", mv);    }    return 0;}

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