UVa 699

题意

先序输入一个带权二叉树, -1即无该儿子

思路

水题
递归输入, 用数组保存每个水平位置的总和即可

一开始没读清楚题, 以为是每一行作为一个二叉树, 本来想用string输入stringstream切割, 观察样例就能推翻

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <sstream>

using namespace std;
const int maxn = 100;
int s[maxn];
int minsize, maxsize;
void build(int,int);

void build( int i ){
    int x;
    cin >> x;
    if( x == -1 )   return;
    s[i] += x;
    minsize = min(minsize, i);
    maxsize = max(maxsize, i);
    build(i-1);
    build(i+1);
}

bool solve( int i ){
    minsize = 101, maxsize = -1;
    memset(s,0,sizeof(s));
    int x;
    cin >> x;
    if( x == -1 )   return false;
    s[i] += x;
    build(i-1);
    build(i+1);
    minsize = min(minsize, i);
    maxsize = max(maxsize, i);
}

int main()
{
    int i, x;
    int casenum = 0;
    while( solve(maxn/2) ){
        for( int j = minsize; j <= maxsize; j++ ){
            if( j == minsize )  cout << "Case " << ++casenum << ':' << endl;
            else    cout << ' ';
            cout << s[j];
        }
        cout << endl << endl;
    }
    return 0;
}
原文地址:https://www.cnblogs.com/JinxiSui/p/9740607.html