刚考完csp，这道题是大模拟题，题意不难理解。以下是题目链接：
http://118.190.20.162/view.page?gpid=T178
当时考场上这道题调了好久没调出来，忽略了很多细节。在这里分享一下满分题解及思路，帮大家避避坑。
#include <iostream>
#include<stdio.h>
#include<queue>
#include<cstring>
#include<string>
#include <map>
#include<cmath>
#include<cstring>
#include<algorithm>
#include <vector>
#include <unordered_set>
#include <unordered_map>
#define MAX 0x3f3f3f3f3f3f //注意这个数要开大一点，不然会溢出
using namespace std;
typedef long long LL;
LL n, m;
// 定义结构体记录每个点的信息，包括其父节点，后代集合，以及其权重
struct point {
int father;
unordered_set<int> sons;
LL weight;
} points[2010];
// 记录每个结点是否被去除
bool used[2010];
// 计算当前剩下的结点的总权重
LL calv(){
LL sum = 0;
for (int i = 1;i <= n;i++) {
if (!used[i]) sum += points[i].weight;
}
return sum;
}
// 用于判断需要测试的点是否是提问点的后代
bool searchfa(int x,int fa) {
if (x == 0) return false;
if (fa == points[x].father || fa == x) return true;
else return searchfa(points[x].father,fa);
}
// 用于构建树
bool fillfa(unordered_set<int> x, int fa) {
if (fa == 0) return false;
unordered_set<int>::iterator it;
for (it = x.begin();it != x.end();it++) {
points[fa].sons.insert(*it);
}
fillfa(x,points[fa].father);
}
// 用于初始化used数组
void fill(int x) {
if(x==0)for (int i = 1;i <= n;i++) used[i]=false;
else for (int i = 1;i <= n;i++) used[i] = true;
}
int main() {
// 输入部分
cin >> n >> m;
for (int i = 1;i <= n;i++) {
cin >> points[i].weight;
}
points[1].father = 0;
for (int i = 2;i <= n;i++) {
int a;
cin >> a;
points[i].father = a;
points[i].sons.insert(i);
fillfa(points[i].sons,a);
points[i].sons.erase(i);
}
// 先求出提问结点，根据测试点是否是其后代分为两类处理
for (int i = 1;i <= m;i++) {
LL target;
LL remain = n;
fill(0);
cin >> target;
// remain记录当前剩余节点数量，为1退出
while (remain != 1) {
LL sum = calv();
int index = 0;
LL w = MAX;
// 求出提问结点index
for (int i = 1;i <= n;i++) {
if(!used[i]) {
unordered_set<int>::iterator it;
LL sum1 = points[i].weight;
for (it = points[i].sons.begin();it != points[i].sons.end();it++) {
if(!used[*it])
sum1 += points[*it].weight;;
}
if (abs(sum - 2 * sum1) < w) {
w = abs(sum - 2 * sum1);
index = i;
}
}
}
cout << index << " ";
// 根据测试点是否是其后代分为两类处理,是后代保留测试点子树
if(searchfa(target,index)){
for (int i = 1;i <= n;i++) {
if(!points[index].sons.count(i))
used[i] = true;
}
used[index] = false;
remain = 1;
unordered_set<int>::iterator it;
for (it = points[index].sons.begin();it != points[index].sons.end();it++) {
if(!used[*it])
remain += 1;
}
}
// 不是后代则去除测试点子树
else{
used[index] = true;
remain -= 1;
unordered_set<int>::iterator it;
for (it = points[index].sons.begin();it != points[index].sons.end();it++) {
if(!used[*it])remain -= 1;
used[*it] = true;
}
}
}
cout << endl;
}
return 0;
}