Promblem I

题意: 给一颗根节点编号为1的树,  节点上都有一个值v[i].  

          每次对树的"根节点为1"的子树进行所有节点都同时加1或减去1的操作,

          求使所有节点为值为0的最小操作次数    

 分析: 从叶子节点开始做操作使所有子叶为零, 

            每次取叶子的加的最大值和减去的最大值对父节点操作,

            如果父节点不为零, 对父节点操作后更新最新的(加减)操作值,此时此父节点变成叶子节点

           直到操作节点规模为0,算法结束

算法:  贪心~, 递归回溯的应用 , 二叉树的遍历

#include <iostream>#include <cstdio>#include <queue>#include <cstring>#include <cmath>#include <vector>#include <set>#include <stack>#include <algorithm>//#include "myAlgorithm.h"#define MAX 100005#define INF (1e17 + 5)#define eps 1e-5using namespace std;typedef long long ll;ll getMax(ll a, ll b){return a < b ? b: a;}vector<vector<int> > adj;int val[MAX];int n;ll ans, add[MAX], red[MAX];void dfs(int node, int father){    int len = adj[node].size();    for(int i = 0; i < len; i++){       int child = adj[node][i];        if(child != father){            dfs(child, node);            add[node] = getMax(add[child], add[node]);            red[node] = getMax(red[child], red[node]);        }    }    val[node] +=add[node] - red[node];    if(val[node] > 0)red[node] +=  val[node];    else add[node] += -val[node];}void init(){    adj.assign(n+ 2, vector<int>() );    int a, b ;    adj[1].push_back(0);    adj[0].push_back(1);    for(int i = 0; i <n - 1; i++){       cin>>a>>b;        adj[a].push_back(b);        adj[b].push_back(a);    }//input end;    add[0] = red[0] = 0;    for(int i = 1; i <= n; i++){        cin>>val[i];        add[i] = 0;        red[i] = 0;    }}int main() {    //freopen("in.txt", "w", stdout);    while(cin>>n){        init();        dfs(1, 0);        cout<<add[1] + red[1]<<endl;        adj.clear();    }    return 0;}