hdu 4714 Tree2cycle
2013-09-09 11:17
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乱搞题,要求用最少的操作把一颗树转化成一个环。其实就是把树分成最少的链,然后连接起来即可,仔细观察树的话会发现,一般一个点的度如果大于1的话,该点必然要断开一些连接,因为最终每个点的度都是2, 然后就是看点上断开那些连接了,其实,如果一个节点的除去父亲节点如果度大于1的话,断开与父亲节点的那条边必然是一种正解。。。于是问题就解决了,只需一遍dfs即可。在杭电交注意加上挂。。。不然会爆栈。。。
#pragma comment(linker,"/STACK:102400000,102400000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#define LL long long
#define PB(a) push_back(a);
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;
const int N = 1111111;
vector<int> E
;
int ans;
int dfs(int u, int fa)
{
int v, i, hav = 0;
for(i = 0; i < E[u].size(); i ++)
{
v = E[u][i];
if(v != fa) hav += dfs(v, u);
}
if(fa == -1 && hav > 1) ans += hav - 2;
else if(hav > 1)
{
ans += hav - 1;
return 0;
}
return 1;
}
int main()
{
int n, t, i, j, u, v;
scanf("%d", &t);
while(t --)
{
scanf("%d", &n);
for(i = 0; i <= n; i ++) E[i].clear();
for(i = 1; i < n; i ++)
{
scanf("%d%d", &u, &v);
E[u].PB(v);
E[v].PB(u);
}
ans = 1;
dfs(1, -1);
printf("%d\n", ans * 2 - 1);
}
return 0;
}
#pragma comment(linker,"/STACK:102400000,102400000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#define LL long long
#define PB(a) push_back(a);
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;
const int N = 1111111;
vector<int> E
;
int ans;
int dfs(int u, int fa)
{
int v, i, hav = 0;
for(i = 0; i < E[u].size(); i ++)
{
v = E[u][i];
if(v != fa) hav += dfs(v, u);
}
if(fa == -1 && hav > 1) ans += hav - 2;
else if(hav > 1)
{
ans += hav - 1;
return 0;
}
return 1;
}
int main()
{
int n, t, i, j, u, v;
scanf("%d", &t);
while(t --)
{
scanf("%d", &n);
for(i = 0; i <= n; i ++) E[i].clear();
for(i = 1; i < n; i ++)
{
scanf("%d%d", &u, &v);
E[u].PB(v);
E[v].PB(u);
}
ans = 1;
dfs(1, -1);
printf("%d\n", ans * 2 - 1);
}
return 0;
}
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