【HDU 5647】DZY Loves Connecting（树DP）
【HDU 5647】DZY Loves Connecting（树DP）
DZY Loves Connecting
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 332 Accepted Submission(s): 112
Total Submission(s): 332 Accepted Submission(s): 112
Problem Description
DZY has an unrooted tree consisting of
n nodes labeled from
1 to
n .
DZY likes connected sets on the tree. A connected set S is a set of nodes, such that every two nodes u,v in S can be connected by a path on the tree, and the path should only contain nodes from S . Obviously, a set consisting of a single node is also considered a connected set.
The size of a connected set is defined by the number of nodes which it contains. DZY wants to know the sum of the sizes of all the connected sets. Can you help him count it?
The answer may be large. Please output modulo 109+7 .
DZY likes connected sets on the tree. A connected set S is a set of nodes, such that every two nodes u,v in S can be connected by a path on the tree, and the path should only contain nodes from S . Obviously, a set consisting of a single node is also considered a connected set.
The size of a connected set is defined by the number of nodes which it contains. DZY wants to know the sum of the sizes of all the connected sets. Can you help him count it?
The answer may be large. Please output modulo 109+7 .
Input
First line contains
t denoting the number of testcases.
t testcases follow. In each testcase, first line contains n . In lines 2∼n , i th line contains pi , meaning there is an edge between node i and node pi . ( 1≤pi≤i−1,2≤i≤n )
( n≥1 ， sum of n in all testcases does not exceed 200000 )
t testcases follow. In each testcase, first line contains n . In lines 2∼n , i th line contains pi , meaning there is an edge between node i and node pi . ( 1≤pi≤i−1,2≤i≤n )
( n≥1 ， sum of n in all testcases does not exceed 200000 )
Output
Output one line for each testcase, modulo
109+7 .
Sample Input
2 1 5 1 2 2 3
Sample Output
1 42HintIn the second sample, the 4 edges are (1,2),(2,3),(2,4),(3,5). All the connected sets are {1},{2},{3},{4},{5},{1,2},{2,3},{2,4},{3,5},{1,2,3},{1,2,4},{2,3,4},{2,3,5},{1,2,3,4},{1,2,3,5},{2,3,4,5},{1,2,3,4,5}. If you need a larger stack size, please use #pragma comment(linker, "/STACK:102400000,102400000") and submit your solution using C++.
Source
BestCoder Round #76 (div.2)
一个挺直接的树DP，之前TC的一个原题，。比赛时死活没啃出来。。。太嫩了……
题目大意：给出一个树，具体建法就不细说了。每个点贡献为1，问树上所有不同集合的合计贡献。
首先对于每个点，我们可以求出它所在的集合数量。
开个数组cnt，表示第i个点在其子树中所在的集合数。这样cnt[i]就等于i所有孩子的cnt+1的乘积。其实这里用到了组合，从所有孩子中可以任意选择一定的集合(+1表示空集)进行组合。
求出集合数是为了求贡献，遍历到i的某个孩子的时候，新增加的贡献其实就是之前的所有贡献乘上该孩子的集合数+1(选择某些集合组合)，然后对于之前遍历的子数，该孩子也可以提供贡献，也就是该孩子的贡献乘上当前出现的集合数。
口述表达的不好，也不会做那种图。。还是看代码吧，就俩公式。。
代码如下：
#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)
using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;
struct Edge
{
int v,next;
};
Edge eg[233333];
int head[233333];
//0存当前节点往下包含当前节点的区间数 1存当前节点的子树中所有包含当前节点的贡献
LL dp[233333][2];
LL ans;
void dfs(int u)
{
dp[u][1] = 1;
dp[u][0] = 1;
for(int i = head[u]; i != -1; i = eg[i].next)
{
dfs(eg[i].v);
//当前点子树中包含当前点的区间的贡献
dp[u][1] = (dp[u][1]*(dp[eg[i].v][0]+1)+dp[eg[i].v][1]*dp[u][0])%mod;
dp[u][0] = (dp[u][0]*(dp[eg[i].v][0]+1))%mod;
}
ans = (ans+dp[u][1])%mod;
}
int main()
{
//fread();
//fwrite();
int x,t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(head,-1,sizeof(head));
for(int i = 2; i <= n; ++i)
{
scanf("%d",&x);
eg[i].v = i;
eg[i].next = head[x];
head[x] = i;
}
ans = 0;
dfs(1);
printf("%lld\n",ans);
}
return 0;
}