【解题报告】Tree Destruction CodeForces - 911F（搜索？？）
【解题报告】Tree Destruction CodeForces - 911F（搜索？？）
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【cf链接】
【vj链接】
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题目
You are given an unweighted tree with n vertices. Then n - 1 following operations are applied to the tree. A single operation consists of the following steps:
- choose two leaves;
- add the length of the simple path between them to the answer;
- remove one of the chosen leaves from the tree.
Initial
Obviously after n - 1 such operations the tree will consist of a single vertex.
Calculate the maximal possible answer you can achieve, and construct a sequence of operations that allows you to achieve this answer!
Input
The first line contains one integer number n (2 ≤ n ≤ 2·105) — the number of vertices in the tree.
Next n - 1 lines describe the edges of the tree in form ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). It is guaranteed that given graph is a tree.
Output
In the first line print one integer number — maximal possible answer.
In the next n - 1 lines print the operations in order of their applying in format ai, bi, ci, where ai, bi — pair of the leaves that are chosen in the current operation (1 ≤ ai, bi ≤ n), ci (1 ≤ ci ≤ n, ci = ai or ci = bi) — choosen leaf that is removed from the tree in the current operation.
See the examples for better understanding.
Examples
input
3
1 2
1 3
output
3
2 3 3
2 1 1
input
5
1 2
1 3
2 4
2 5
output
9
3 5 5
4 3 3
4 1 1
4 2 2
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题目大意
有棵树，按照某规律删除结点并得到一个值。
某规律：找俩点，删掉其中一个
一个值：每次操作找的那俩点的距离
样例输入
第一行：一个n，表示树上有几个点
第2-n行：两个数x,y，表示一条边
输出
第一行：刚刚说的那一个值
第2-n行：三个数x,y,z，x,y表示一条边，z表示删除的那个点
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解题思路
如果已经有一条路径（直径）是最长的，那其他点的最长路径一定是到达直径的端点的
假设这两个点叫p1,p2
要求的一个值叫ans
- 步骤一：找到最长路径（两次dfs)
（从一个点出发，找到它的最长路径，那另一个点一定属于直径的端点）
- 步骤二：除最长路径上的点全部删掉
- 步骤三：删掉直径
（步骤二三通过分别从p1,p2遍历得到长度dis[i],dis2[i]，用d[i]表示在直径上的某点离p2的距离，然后遍历一次，回溯时删除）
（求ans的过程在遍历之前，直接加d[i]或dis[i]或dis2[i]即可）
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AC代码
不建议看这个代码，太乱了，幸好是1A，不然都不知道怎么改……
#include if(!vis[edge[x][i]])
{
vis[edge[x][i]]=1;
dis[edge[x][i]]=dis[x]+1;
dfs1(edge[x][i]);
}
}
}
void dfs3(int x)//求x到各点的距离存在dis2数组里
{
for(int i=0;iif(!vis[edge[x][i]])
{
vis[edge[x][i]]=1;
dis2[edge[x][i]]=dis2[x]+1;
dfs3(edge[x][i]);
}
}
}
int dfs2(int x)//求x到直径上各点的距离，存在d数组里
{
if(x==p2) return d[x]=1;
for(int i=0;iif(!vis[edge[x][i]])
{
vis[edge[x][i]]=1;
if(dfs2(edge[x][i]))
{
d[x]=d[edge[x][i]]+1;
return 1;
}
}
}
return 0;
}
int dfs4(int x)//输出除直径外的点
{//printf("A");
for(int i=0;i//printf("B");
if(!vis[edge[x][i]])
{
vis[edge[x][i]]=1;
dfs4(edge[x][i]);
if(d[edge[x][i]]==0)
{//printf("D");
if(dis[edge[x][i]]>=dis2[edge[x][i]]) printf("%d %d %d\n",p1,edge[x][i],edge[x][i]);
else printf("%d %d %d\n",p2,edge[x][i],edge[x][i]);
}
}
}
}
int main()
{
int n,i,j,x,y,maxlen;
scanf("%d",&n);
for(i=1;iscanf("%d%d",&x,&y);
edge[x].push_back(y);
edge[y].push_back(x);
}
vis[1]=1;
dis[1]=0;
dfs1(1);
maxlen=-1;
for(i=1;i<=n;i++)
{
if(maxlen0;
}
vis[p1]=1;
dis[p1]=0;
dfs1(p1);
maxlen=-1;
for(i=1;i<=n;i++)
{
if(maxlen0;
}
//printf("p1=%d p2=%d \n",p1,p2);
dfs2(p1);
dis2[p2]=0;
for(i=1;i<=n;i++)
vis[i]=0;
vis[p2]=1;
dfs3(p2);
LL ans=0;
for(i=1;i<=n;i++)
{
//printf("%d %d %d\n",dis[i],dis2[i],d[i]);
if(!d[i]) ans+=max(dis[i],dis2[i]);
else ans+=d[i]-1,print[d[i]]=i;
}
printf("%lld\n",ans);
for(i=1;i<=n;i++)
vis[i]=0;
vis[p1]=1;
dfs4(p1);
for(i=1;iprintf("%d %d %d\n",p1,print[i],print[i]);//print记录的是那条直径
}
return 0;
}