(树形DP) acdream 1028

Path

Time Limit: 4000/2000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)
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Problem Description

Check if there exists a path of length l in the given tree with weight assigned to each edges.

Input

The first line contains two integers n and q, which denote the number of nodes and queries, repectively.

The following (n−1) with three integers ai,bi,ci, which denote the edge between ai and bi, with weight ci.

Note that the nodes are labled by 1,2,…,n.

The last line contains q integers l1,l2,…,lq, denote the queries.

(1≤n,q≤105,1≤ci≤2)

Output

For each query, print the result in seperated line. If there exists path of given length, print "Yes". Otherwise, print "No".

Sample Input

4 6
1 2 2
2 3 1
3 4 2
0 1 2 3 4 5

Sample Output

Yes
Yes
Yes
Yes
No
Yes

Source

ftiasch

Manager

nanae

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<vector>
#define INF 100000000
using namespace std;
vector<int> e[100005],w[100005];
int n,q;
int dp[100005][2],val[2];
void dfs(int u,int father)
{
dp[u][0]=0;
dp[u][1]=-INF;
for(int i=0;i<e[u].size();i++)
{
int v=e[u][i];
int ww=w[u][i];
if(v==father)
continue;
dfs(v,u);
for(int x=0;x<2;x++)
{
for(int y=0;y<2;y++)
{
val[(x+y+ww)&1]=max(val[(x+y+ww)&1],dp[u][x]+dp[v][y]+ww);
}
}
for(int x=0;x<2;x++)
dp[u][(x+ww)&1]=max(dp[u][(x+ww)&1],dp[v][x]+ww);
}
}
int main()
{
scanf("%d%d",&n,&q);
for(int i=1;i<n;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
e[x].push_back(y);
e[y].push_back(x);
w[x].push_back(z);
w[y].push_back(z);
}
val[0]=val[1]=-INF;
dfs(1,-1);
while(q--)
{
int x;
scanf("%d",&x);
if(x<0)
printf("No\n");
else
{
if(x<=val[x&1])
printf("Yes\n");
else
printf("No\n");
}
}
return 0;
}