题目描述
We have a tree with N vertices. Vertex 1 is the root of the tree, and the parent of Vertex i (2≤i≤N) is Vertex Pi.
To each vertex in the tree, Snuke will allocate a color, either black or white, and a non-negative integer weight.
Snuke has a favorite integer sequence, X1,X2,…,XN, so he wants to allocate colors and weights so that the following condition is satisfied for all v.
The total weight of the vertices with the same color as v among the vertices contained in the subtree whose root is v, is Xv.
Here, the subtree whose root is v is the tree consisting of Vertex v and all of its descendants.
Determine whether it is possible to allocate colors and weights in this way.
Constraints
1≤N≤1 000
1≤Pi≤i−1
0≤Xi≤5 000
输入
Input is given from Standard Input in the following format:
N
P2 P3 … PN
X1 X2 … XN
输出
If it is possible to allocate colors and weights to the vertices so that the condition is satisfied, print POSSIBLE; otherwise, print IMPOSSIBLE.
样例输入
3
1 1
4 3 2样例输出
POSSIBLE提示
For example, the following allocation satisfies the condition:
Set the color of Vertex 1 to white and its weight to 2.
Set the color of Vertex 2 to black and its weight to 3.
Set the color of Vertex 3 to white and its weight to 2.
There are also other possible allocations.如下图，粉色为x[i]，蓝色为(f[i],x[i])。f[i]代表 以i为跟的所有的 与i结点颜色不同的 子节点 的权值和。
显然，对每个节点，咱们应该尽可能的用子节点的权值将本结点的x[i]填满，这就成了背包问题了。在这里，我们对结点i 做dp的时候，结点i的每个孩子结点j(不包含孙子结点）中，必须在f[j]和x[j]中选择一个，因为如果i和j不同色，那i就必须选择与j不同色的 j的子节点 的权值和，即f[j];否则就选择x[j]。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,p;
struct edge{
ll en,next;ll cost;
}e[300010];
int w[1010],x[1010];
int head[300010],cnt=0,vis[200010]={0};
void add(ll start,ll ennd,ll c){//头插法，单向
e[cnt].en=ennd;
e[cnt].next=head[start];
e[cnt].cost=c;
head[start]=cnt;//记录下标
cnt++;
}
int flag=1,dp[5050][5050],f[5050],W=0,b=0;
int dfs(int u){
if(vis[u]||flag==0) return 0;
vis[u]=1;
f[u]=0x3f3f3f3f;
memset(dp[u],0x3f3f3f3f,sizeof(dp[u]));
dp[u][0]=0;
if(head[u]==-1) {f[u]=0;return 0;}
for(int i=head[u];i!=-1;i=e[i].next){
dfs(e[i].en);
for(int j=x[u];j>=0;j--){
int temp=0x3f3f3f3f;
if(j-x[e[i].en]>=0) temp=min(temp,dp[u][j-x[e[i].en]]+f[e[i].en]);
if(j-f[e[i].en]>=0) temp=min(temp,dp[u][j-f[e[i].en]]+x[e[i].en]);
dp[u][j]=temp;
}
}
for(int i=0;i<=x[u];i++)
f[u]=min(f[u],dp[u][i]);
if(f[u]==0x3f3f3f3f){ flag=0;return 0;}
}
int main(){
scanf("%d",&n);
for(int i=0;i<=n;i++){
head[i]=-1;
}
for(int i=2;i<=n;i++){
scanf("%d",&p);
add(p,i,1);
}
for(int i=1;i<=n;i++)
{
scanf("%d",&x[i]);
}
dfs(1);
if(flag){
printf("POSSIBLE\n");
}
else{
printf("IMPOSSIBLE\n");
}
return 0;
}