(树形DP) acdream 1028

Path

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

Submit Statistic Next Problem

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;
}