紫书 The Falling Leaves UVA - 699 递归得简单

Posted 2020-10-30 COLORFUL

题意：给你一颗二叉树的前序遍历，空子树以-1表示，将左右子树的权值投影到一维数轴上，左儿子位置为根位置-1，右儿子+1求个个整点上的和；

题解：递归，整个过程只需维护一个sum数组。

　　　　更新根，更新leftson ,更新rightson;

代码：

#define _CRT_SECURE_NO_WARNINGS
#include "stdio.h"
#include<stdio.h>
#include<algorithm>
#include<string>
#include<vector>
#include<list>
#include<set>
#include<iostream>
#include<string.h>
#include<queue>
#include<string>
#include<sstream>
using namespace std;
const int maxn = 80+5;
int sum[maxn],p;
void build(int p) {//更新以p为根的树对sum的贡献：更新根，更新leftson ,更新right son;
    int v; cin >> v;
    if (v == -1)return;
    sum[p] += v;
    build(p - 1); build(p + 1);
}
bool init() {
    int v;
    cin >> v;
    if (v == -1)return false;
    memset(sum, 0, sizeof(sum));
    int pos = maxn / 2;
    sum[pos] = v;
    build(pos - 1); build(pos + 1);
}
int main(){
    int kase = 0;
    while (init()) {
        int p = 0;
        while (sum[p] == 0)p++;
        cout << "Case " << ++kase << ":\n" << sum[p++];
        while (sum[p] != 0)cout << " " << sum[p++];
        cout << "\n\n";
    }
    //system("pause");
    return 0;
}