传送门
很基础的一道树形dp,就是输入比较恶心一点.
假设树的跟节点是0号节点,用dp[i][0]和dp[i][1]来表示i号节点放哨兵和不放哨兵的mincost.转移方程(y为i的儿子节点):
dp[i][1] = 求和{min(dp[y][0],dp[y][1])};
dp[i][0] = 求和{dp[y][1]};
因为这道题是要看到边不是看到点,所以一条边对应的两个点一定要放至少一个哨兵.还是比较简单的.另外一题是要看点,比较麻烦一点.
代码
#pragma GCC optimize(2)
#define LL long long
#define pq priority_queue
#define ULL unsigned long long
#define pb push_back
#define mem(a,x) memset(a,x,sizeof a)
#define pii pair<int,int>
#define fir(i,a,b) for(int i=a;i<=(int)b;++i)
#define afir(i,a,b) for(int i=(int)a;i>=b;--i)
#define ft first
#define vi vector<int>
#define sd second
#define ALL(a) a.begin(),a.end()
#define bug puts("-------")
#define mpr(a,b) make_pair(a,b)
#include <bits/stdc++.h>
using namespace std;
const int N = 3e3+10;
inline void read(int &a){
int x = 0,f=1;char ch = getchar();
while(ch<'0'||ch>'9'){
if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){
x=x*10+ch-'0';ch=getchar();}
a = x*f;
}
int nxt[N],head[N],ct=1,to[N],dp[N][2];
void addedge(int x,int y){
nxt[++ct] = head[x];head[x] = ct;to[ct] = y;
}
void init(){
mem(head,0);
mem(dp,0);
ct = 1;
}
void dfs(int u,int fa){
dp[u][1] = 1;
for(int i = head[u];i;i=nxt[i]){
int y = to[i];
if(y == fa) continue;
dfs(y,u);
dp[u][1] += min(dp[y][1],dp[y][0]);
dp[u][0] += dp[y][1];
}
}
int main(){
int n;
while(cin >> n){
init();
string s;
fir(i,1,n){
cin >> s;
int x=0,nodenum=0,idx=0;
while(isdigit(s[idx++])) x = x*10+s[idx-1]-'0';
idx += 1;
while(isdigit(s[idx++])) nodenum = nodenum*10 + s[idx-1]-'0';
fir(j,1,nodenum){
int y;
cin >> y;
addedge(x,y);
addedge(y,x);
}
}
dfs(0,-1);
cout << min(dp[0][1],dp[0][0]) << endl;
}
return 0;
}