NBUT 1186 Get the Width（DFS求树的宽度，水题）

• [1186] Get the Width

• 时间限制: 1000 ms　内存限制: 65535 K
• 问题描述
• It‘s an easy problem. I will give you a binary tree. You just need to tell me the width of the binary tree.
  The width of a binary tree means the maximum number of nodes in a same level.

  

  For example, the width of binary tree above is 3.

• 输入
• The first line is an integer T, means the number of cases.
  Then follow T cases.
  For each case, the first line is an integer N, means the number of nodes. (1 <= N <= 10)
  Then follow N lines. Each line contains 3 integers P A B; indicate the number of this node and its two children node. If the node doesn’t have left child or right child, then replace it by -1.
  You can assume the root is 1.

• 输出
• For each case, output the width.
• 样例输入

• 1
  6
  4 -1 -1
  2 4 5
  5 -1 -1
  1 2 3
  6 -1 -1
  3 -1 6
  

• 样例输出

• 3

题目链接：NBUT 1186

闲来无事水一发简单的= =，听说大二要学数据结构，原来树的宽度是这么个意思，建图dfs一下即可

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=100010;
struct edge
{
    int to;
    int pre;
};
edge E[N];
int head[N],tot;
int cnt[N],vis[N];
void add(int s,int t)
{
    E[tot].to=t;
    E[tot].pre=head[s];
    head[s]=tot++;
}
void init()
{
    CLR(cnt,0);
    CLR(head,-1);
    tot=0;
    CLR(vis,0);
}
void dfs(int cur,int dep)
{
    ++cnt[dep];
    vis[cur]=1;
    for (int i=head[cur]; ~i; i=E[i].pre)
    {
        int son=E[i].to;
        if(!vis[son])
            dfs(son,dep+1);
    }
}
int main(void)
{
    int tcase,i,j,n,p,a,b;
    scanf("%d",&tcase);
    while (tcase--)
    {
        init();
        scanf("%d",&n);
        while (n--)
        {
            scanf("%d%d%d",&p,&a,&b);
            if(a!=-1)
                add(p,a);
            if(b!=-1)
                add(p,b);
        }
        dfs(1,1);
        printf("%d\n",*max_element(cnt+1,cnt+N));
    }
    return 0;
}