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Saber is a Class in the Holy Grail War. This Class is regarded as one of the most powerful Class.
Saber is a tree-lover. She loves all kinds of trees. One day, she suddenly comes up with a curious problem. She wants to know that in the path between x and y, whether it exists that when we choose three different edges i,j,k, the length of these three edges can build a triangle. Now she wants you to solve this problem.
Input
There are multiple test cases. The first line of input contains an integer T(T ≤ 5), indicating the number of test cases. For each test case:
The first line contains one integer N(1 ≤ N ≤ 100000), indicating the number of tree’s vertices. In the following N-1 lines, there are three integers x, y, w (1 ≤ w ≤ 1000000000), indicating an edge weighted w between x and y.
In the following line also contains one integer Q(1 ≤ Q ≤ 100000), indicating the number of queries. In the following Q lines, there are two integers x, y, indicating a query between x and y.
Output
For each test case output ‘Case #i:’ in the first line, i equals to the case number. Then for every query output ‘Yes’ or ‘No’ in one line.
Sample Input
2 5 1 2 5 1 3 20 2 4 30 4 5 15 2 3 4 3 5 5 1 4 32 2 3 100 3 5 45 4 5 60 2 1 4 1 3
Sample Output
Case #1: No Yes Case #2: No Yes
解题思路:
1、因为题目说明了是树,所以保证给出的数据是一定能通的。2、存在的路径中找到是否存在三条边构成三角形,此时的路径是两点的唯一路径,不包括其他死胡同的边(哪怕是相通的)。3、斐波那契数列与三角形的关联(破 TLE)。
AC 代码(DFS 版)
#include<bits/stdc++.h>
#include<cmath>
#define mem(a,b) memset(a,b,sizeof a);
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn=100000+100;
struct node
{
int e,w;
//node(int e,int w):e(e),w(w){} // 推荐这种写法,用到时,无需再定义声明
};
vector<node> ve[maxn];
int vis[maxn], num[60];
int n,m,s,e,flag,found;
void init()
{
for(int i=0;i<=maxn;i++)
// ve[i].clear();
vector<node>().swap(ve[i]); // 这种写法比 clear() 更彻底清除
m=n-1;
mem(vis,0);
}
int ok(int a,int b,int c)
{
return a+b>c;
}
void dfs(int s,int step)
{
// 50的标记是来自于斐波那契数列上限46就达到了此题目中的最大值
// 所以顶峰是45时,再无论加一条边就肯定能找到构成的三角形的三条边
if(step>50 || found) return;
if(s==e)
{
found=1;
if(step>50)
{
flag=1;
return;
}
sort(num,num+step);
int a,b,c;
for(int i=2;i<step;i++) // 判断是否可以构成三角形
{
a=num[i-2], b=num[i-1], c=num[i];
if(ok(a,b,c))
{
flag=1;
return;
}
}
return;
}
for(int i=0;i<ve[s].size();i++)
{
if(!vis[ve[s][i].e])
{
vis[ve[s][i].e]=1;
num[step]=ve[s][i].w;
dfs(ve[s][i].e,step+1);
vis[ve[s][i].e]=0;
}
}
return;
}
int main()
{
int T,kase=0; scanf("%d",&T);
while(T-- && ~scanf("%d",&n))
{
init();
int u,v,w,q;
node nd1,nd2;
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
nd1.e=v; nd1.w=w;
nd2.e=u; nd2.w=w;
ve[u].push_back(nd1);
ve[v].push_back(nd2);
}
scanf("%d",&q);
printf("Case #%d:\n",++kase);
for(int i=0;i<q;i++)
{
scanf("%d%d",&s,&e);
found=flag=0;
vis[s]=1; // 别忘记一开始就已经经过的这个点s
dfs(s,0);
vis[s]=0;
// found == 0 是因为在到终点的途中就已经超过了阈值,所以就可以马上return,此时的found一定等于0
puts(!found || flag?"Yes":"No");
}
}
return 0;
}