6481: Integers on a Tree

最新推荐文章于 2022-10-18 23:57:07 发布

Acerkoo

最新推荐文章于 2022-10-18 23:57:07 发布

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6481: Integers on a Tree

时间限制: 1 Sec 内存限制: 128 MB
提交: 69 解决: 12
[ 提交][ 状态][ 讨论版][命题人: admin]

题目描述

We have a tree with N vertices. The vertices are numbered 1,2,…,N. The i-th (1≤i≤N−1) edge connects the two vertices Ai and Bi.
Takahashi wrote integers into K of the vertices. Specifically, for each 1≤j≤K, he wrote the integer Pj into vertex Vj. The remaining vertices are left empty. After that, he got tired and fell asleep.
Then, Aoki appeared. He is trying to surprise Takahashi by writing integers into all empty vertices so that the following condition is satisfied:
Condition: For any two vertices directly connected by an edge, the integers written into these vertices differ by exactly 1.
Determine if it is possible to write integers into all empty vertices so that the condition is satisfied. If the answer is positive, find one specific way to satisfy the condition.

Constraints
1≤N≤105
1≤K≤N
1≤Ai,Bi≤N (1≤i≤N−1)
1≤Vj≤N (1≤j≤K) (21:18, a mistake in this constraint was corrected)
0≤Pj≤105 (1≤j≤K)
The given graph is a tree.
All vj are distinct.

输入

The input is given from Standard Input in the following format:
N
A1 B1
A2 B2
:
AN−1 BN−1
K
V1 P1
V2 P2
:
VK PK

输出

If it is possible to write integers into all empty vertices so that the condition is satisfied, print Yes. Otherwise, print No.

样例输入

样例输出

Yes

提示

The figure below shows the tree when Takahashi fell asleep. For each vertex, the integer written beside it represents the index of the vertex, and the integer written into the vertex is the integer written by Takahashi.

Aoki can, for example, satisfy the condition by writing integers into the remaining vertices as follows:

来源

ABC047&ARC063

题意：

建了一棵树，在m个位置上标记一个数，判断是否能形成一个有边连接的两个节点之间的数值只差1 。

思路：

对每个节点而言，其上下限的值一定由其父想上下分别扩大 1 ，并且奇偶性与其父节点想反，那么我们维护每个节点的上下限值，父节点求其所有子节点的上下限的交集。

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10,inf=1e9+7;
int l[maxn],r[maxn],n,m,k,t;
bool flag=true;
vector<int> e[maxn];
void dfs(int x,int fa=0){
    for (auto u:e[x]){
        if(u==fa) continue;
        dfs(u,x);
//        printf("~%d %d\n",x,u);
        if(r[u]==inf) continue;
        l[u]--,r[u]++;
        if(r[x]==inf){
            l[x]=l[u],r[x]=r[u];
            continue;
        }
        if((r[x]^r[u])&1){
            flag=false;
            continue;
        }
        l[x]=max(l[x],l[u]);
        r[x]=min(r[x],r[u]);
    }
    if(l[x]>r[x]) flag=false;
}

int main(){
    scanf("%d",&n);
    for (int i=1,u,v; i<n; i++){
        scanf("%d%d",&u,&v);
        e[u].push_back(v);
        e[v].push_back(u);
    }
    for (int i=1;i<=n; i++) l[i]=-inf,r[i]=inf;
    scanf("%d",&m);
    int x,y;
    while(m--){
        scanf("%d%d",&x,&y);
        l[x]=r[x]=y;
    }
    dfs(1);
    puts(flag?"Yes":"No");
    return 0;
}