树上dfs + 思维
2017-02-21 20:20
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the number lamp ai,
it is hanging on (and 0, if is there is no such lamp), 表示的是祖先的序号
链接:点击打开链接
题目中的坑:应该自下而上的判断子树的累加和是否满足条件。如果是自上而下的判断,一旦该树的子树有满足条件的,去掉子树后又满足条件这种情况是无法识别出来。只能一回溯,且实际上回溯的复杂度并不高,相当于遍历所有点两遍。
解析:dfs 判断累加和。如果一个子树满足条件,价值变为零,两棵及以上满足条件则成功(恰好两棵需要额外判断不是根节点)
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int const maxn = 1000006;
vector<int>son[maxn];
int t[maxn],p,root;
int pa[maxn],ans[2];
int num = 0;
void dfs(int x)
{
if(num == 2)return;
int temp;
for(int i = 0; i < son[x].size(); i ++)
{
temp = son[x][i];
dfs(temp);
t[x] += t[temp];
}
if(t[x] == p && x != root)
{
ans[num] = x;
num ++;
t[x] = 0;
if(num == 2)return;
}
}
int main()
{
// freopen("in.txt","r",stdin);
int n;
int sum = 0;
scanf("%d",&n);
for(int i = 1; i <= n; i ++)
{
scanf("%d%I64d",&pa[i],&t[i]);
sum += t[i];
son[pa[i]].push_back(i);
if(pa[i] == 0)root = i;
}
bool flag = false;
if(sum % 3 != 0 || sum == 0)
{
flag = true;
}
else
{
p = sum / 3;
dfs(root);
if(num >= 2)
{
printf("%d %d\n",ans[0],ans[1]);
}
else
{
flag = true;
}
}
if(flag == true)
printf("-1\n");
return 0;
}
it is hanging on (and 0, if is there is no such lamp), 表示的是祖先的序号
链接:点击打开链接
题目中的坑:应该自下而上的判断子树的累加和是否满足条件。如果是自上而下的判断,一旦该树的子树有满足条件的,去掉子树后又满足条件这种情况是无法识别出来。只能一回溯,且实际上回溯的复杂度并不高,相当于遍历所有点两遍。
解析:dfs 判断累加和。如果一个子树满足条件,价值变为零,两棵及以上满足条件则成功(恰好两棵需要额外判断不是根节点)
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int const maxn = 1000006;
vector<int>son[maxn];
int t[maxn],p,root;
int pa[maxn],ans[2];
int num = 0;
void dfs(int x)
{
if(num == 2)return;
int temp;
for(int i = 0; i < son[x].size(); i ++)
{
temp = son[x][i];
dfs(temp);
t[x] += t[temp];
}
if(t[x] == p && x != root)
{
ans[num] = x;
num ++;
t[x] = 0;
if(num == 2)return;
}
}
int main()
{
// freopen("in.txt","r",stdin);
int n;
int sum = 0;
scanf("%d",&n);
for(int i = 1; i <= n; i ++)
{
scanf("%d%I64d",&pa[i],&t[i]);
sum += t[i];
son[pa[i]].push_back(i);
if(pa[i] == 0)root = i;
}
bool flag = false;
if(sum % 3 != 0 || sum == 0)
{
flag = true;
}
else
{
p = sum / 3;
dfs(root);
if(num >= 2)
{
printf("%d %d\n",ans[0],ans[1]);
}
else
{
flag = true;
}
}
if(flag == true)
printf("-1\n");
return 0;
}
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