【bzoj1912】[Apio2010] patrol 巡逻 树形dp

题目大意：给你一棵树，你可以连k条边(1<=k<=2)，使得每个点都遍历过一次且经过的边数最少。

题目分析：对于一棵树来说，遍历所有的点最少需要将每条边走两次，即2*（n-1）。显然当k==1时将直径的两个端点相连，便可以使原来直径上的边减少1次，显然最优。当k==2时，只要将原来直径上的边权赋为-1再找一次树的直径即可。最终答案ans=2*（n-1）-dis1-dis2+2;

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#include<cmath>#include<cctype>#include<cassert>#include<climits>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForD(i,n) for(int i=n;i;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define RepD(i,n) for(int i=n;i>=0;i--)#define MEM(a) memset(a,0,sizeof(a))#define MEMI(a) memset(a,127,sizeof(a))#define MEMi(a) memset(a,128,sizeof(a))#define INF (2139062143)#define phiF (1000000006)#define MAXN (1000000+10)typedef long long LL;struct info{int to,next,val;}e[200005];int tot=1,first[100005],n,k,x,y,las1[100005],las2[100005],dis,last;void add(int x,int y){    tot++;e[tot].to=y;e[tot].next=first[x];e[tot].val=1;first[x]=tot;}int dfs(int u,int fa){int max1=0,max2=0;for (int p=first[u];p;p=e[p].next){int v=e[p].to;if (v!=fa){    int tmp;tmp=e[p].val+dfs(v,u);if (tmp>max1) max2=max1,max1=tmp,las2[u]=las1[u],las1[u]=p;else if (tmp>max2)  max2=tmp,las2[u]=p;    }}if (max1+max2>dis) dis=max1+max2,last=u;return max1;}int main(){scanf("%d%d",&n,&k);For (i,n-1){scanf("%d%d",&x,&y);add(x,y);add(y,x);}dfs(1,0);int ans=2*(n-1);    ans-=dis-1;if (k==2){dis=0;for (int p=las1[last];p;p=las1[e[p].to]) e[p].val=e[p^1].val= -1;for (int p=las2[last];p;p=las1[e[p].to]) e[p].val=e[p^1].val= -1;dfs(1,0);    ans-=dis-1;}printf("%d",ans);}