ACdream 1028 Path

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那么给出一条路径，长度为X，(X>2)，我们考虑他的两端，可能全为1，可能也有2，如果把边权为2的边删掉，或者把两端边权为1的都删了，那么必然存在一条长度为X-2的边。

如果 x 是可行的，那么 (x−2) 也是可行的，所以可以动态规划求出最长的奇数和偶数。注意负数的情况。

Path

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

SubmitStatisticNext 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 61 2 22 3 13 4 20 1 2 3 4 5

Sample Output

YesYesYesYesNoYes

Source

ftiasch

Manager

nanae

SubmitStatistic

const int inf = 0x3f3f3f3f;const int N = 201010;int n, Q;int fa[N];int f(int x){    return x==fa[x] ? x : fa[x] = f(fa[x]);}#define UPD(x, a) x=max(x,a);int to[N], head[N], cost[N], next[N];int en;void init_edge(){    en = 0;    memset(head, -1, sizeof head);    for(int i=1; i<=n; i++) fa[i] = i;}void add(int u, int v, int w){    to[++en] = v,    cost[en] = w;    next[en] = head[u];    head[u] = en;}int down[N][2];int lim[N];void dfs(int u, int p){    down[u][0] = 0;    down[u][1] = -inf;    for(int i=head[u]; ~i; i=next[i])    {        int v = to[i], w = cost[i];        if(v == p) continue;        dfs(v, u);        for(int x = 0; x<2; x++)            for(int y=0; y<2; y++)            {                UPD(lim[x+y+w & 1], down[u][x] + down[v][y] + w);            }        for(int y=0; y<2; y++)            UPD(down[u][y+w & 1], down[v][y] + w);    }}int main(){    while (cin >> n >>Q)    {        init_edge();        for (int i=1; i<n; i++)        {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            add(u, v, w);            add(v, u, w);            fa[f(u)] = f(v);        }        lim[0] = lim[1] = -inf;        dfs(1, -1);        while (Q --)        {            int l;            scanf("%d", &l);            if (l < 0)                puts("No");            else            {                puts(l <= lim[l%2] ? "Yes" : "No");            }        }    }}

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