[Dp] && solution to a problem tree caterpillar Luo Gu P3174

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Subject description:

For a tree, we can be a strand and the strand sides connected to the out come into looks like a caterpillar, the more points, the greater the caterpillar. Such as trees left figure (Fig. 1) extracting a portion of the right side becomes a caterpillar (FIG. 2).

Input format
of the first row two integers N, M, respectively, the number of edges and the number of nodes in the tree, the tree.
Next M rows, each row two integers a, b represent points a and b have connecting edges (a, b≤N). You can not assume that a pair of identical (a, b) appears more than once.

Output formats
An integer representing the size of the largest caterpillars.

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analysis:

Questions can be converted to make thee a chain, the maximum number of nodes contained in this chain.
For a node x, we $Set dp [x] is the root of x is the maximum number of nodes$

We updated by enumerating the child node x y
difficult to find, for every x, y at most only choose one, while the other child nodes also be included, because directly connected with him.
$Provided NUM [x] denotes the point x the number connected$

Then we can get the following equation:
$dp[x]=max(dp[y]+num[x]-1)\ (y\in son(x))$

So the demand equation dp Well, consider how to find the answer.

Not difficult to find, in fact, the final answer $The number of times with the longest chain length of the chain and, together with the nodes -1$。

We only need to update the answer to when dp.

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Code

#include<bits/stdc++.h>
using namespace std;
struct node{
    int y,Next;
}e[700000];
int linkk[700000];
int sum[700000];
int num[700000];
int dp[700000];
int n,m,len=0;
//int maxx1;
//int maxx2;
int maxxx=0;
void insert(int x,int y){
    e[++len].Next=linkk[x];
    e[len].y=y;
    linkk[x]=len;
}
void find_sum(int x,int fa){
    sum[x]=1;
    for (int i=linkk[x];i;i=e[i].Next){
	    int y=e[i].y;
	    if (y==fa) continue;
	    find_sum(y,x);
	    sum[x]+=sum[y];
	}
}
void treedp(int x,int fa){
	int maxx1=0,maxx2=0;
	dp[x]=1;
    for (int i=linkk[x];i;i=e[i].Next){
	    int y=e[i].y;
	    if (y==fa) continue;
	    treedp(y,x);
	    if (dp[y]>maxx1)
	      maxx2=maxx1,maxx1=dp[y];
	    else maxx2=max(maxx2,dp[y]);//更新最长链和次长链
	    dp[x]=max(dp[x],dp[y]+num[x]-1);//更新
	}
	maxxx=max(maxxx,maxx1+maxx2+num[x]-1);//更新
}
int main(){
	freopen("WORM.in","r",stdin);
	freopen("WORM.out","w",stdout);
	scanf("%d %d",&n,&m);
	for (int i=1,x,y;i<=m;i++)
	  scanf("%d %d",&x,&y),insert(x,y),insert(y,x),num[x]++,num[y]++;//记录相连的点的个数
	for (int i=1;i<=n;i++) dp[i]=1;
	find_sum(1,0);//突然发现没什么用。。、。
	treedp(1,0);
	printf("%d",maxxx);
	return 0;
}