Gym-101353H Simple Path（树型dp）

最新推荐文章于 2020-11-27 13:53:11 发布

PinkAir

最新推荐文章于 2020-11-27 13:53:11 发布

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分类专栏： ACM

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本文探讨了一种在加权根节点树中计算所有简单路径长度之和的方法，并通过模运算输出结果。重点在于理解如何通过深度优先搜索（DFS）遍历子树并累积路径长度。

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H. Simple Path

Score: 100

CPU: 1s

Memory: 1024MB

You will be given a weighted rooted tree, where root is node 1. For each subtree of that tree, you will have to compute the summation of lengths of all possible simple paths present in that subtree. Eventually you need to print the summation of those values modulo 1000000007.

Image of the tree in the first sample is given below.

Input Specification:

First line of input will contain an integer T (1 <= T <= 10) denoting the number of test cases. Each test case will begin with an integer N, indicating number of nodes. The following N-1 lines will have three integers each, U (1 <= U <= N), V (1 <= V <= N), W (1 <= W <= 1000000000), which denotes that there is an edge in the tree from node U to node V having W weight.

Input File is large. Use fast I/O methods.

Subtask 1: 1<= N <=1000 (15 points)

Subtask 2: 1<= N <=100000 (85 points)

Output Specification For each test case, print the case number, and then print the answer to the problem modulo 1000000007 in separate lines.

Sample

Input

1 2 3

1 3 2

2 4 1

2 5 4

3 6 6

3 7 8

1 2 3

1 3 2

1 4 4

3 5 7

3 6 1

Output

Case 1: 212

Case 2: 109

Explanation of the 1st sample:

Summation of lengths of all possible sample paths for each subtree is given below:

1: 174

2: 10

3: 28

4: 0

5: 0

6: 0

7: 0

Explanation of the 2nd sample: Summation of lengths of all possible sample paths for each subtree is given below:

1: 93

2: 0

3: 16

4: 0

5: 0

6: 0

c[i]是每个子树的节点个数 a[i]是i与每个子节点距离之和 sum[i]是i这颗树的价值

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int ,int > P;
#define INF 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define mem(a) memset(a,0,sizeof(a))
const int Max=200000+10;
const int mod=1000000007;
int t,n;
ll a[Max],c[Max],sum[Max];
struct node {
	int to,dis;
};
vector<node> g[Max];
void dfs(int x,int fa) {
	c[x]=1;
	for(auto &i:g[x]) {
		if(i.to==fa) continue;
		dfs(i.to,x);
		c[x]+=c[i.to];
		a[x]=(a[x]+a[i.to]+c[i.to]*i.dis%mod)%mod;
	}

	for(auto &i:g[x]) {
		if(i.to==fa) continue;
		sum[x]=(sum[x]+(c[x]-c[i.to])*(a[i.to]+c[i.to]*i.dis%mod)+sum[i.to])%mod;
	}
}
int main() {
	scanf("%d",&t);
	int ans=0;
	while(t--) {
		scanf("%d",&n);
		for(int i=1; i<=n; i++)
			g[i].clear();

		int u,v,w;
		for(int i=1; i<n; i++) {
			scanf("%d%d%d",&u,&v,&w);
			g[u].pb(node {v,w});
			g[v].pb(node {u,w});
		}
		mem(a);
		mem(sum);
		dfs(1,0);
		ll temp=0;
		for(int i=1; i<=n; i++)
			temp=(temp%mod+sum[i]%mod)%mod;
		printf("Case %d: %lld\n",++ans,temp);
	}
	return 0;
}