Rebuilding Roads
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 10066 | Accepted: 4595 |
Description
The cows have reconstructed Farmer John‘s farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The cows didn‘t have time to rebuild any extra roads, so now there is exactly one way to get from any given barn to any other
barn. Thus, the farm transportation system can be represented as a tree.
Farmer John wants to know how much damage another earthquake could do. He wants to know the minimum number of roads whose destruction would isolate a subtree of exactly P (1 <= P <= N) barns from the rest of the barns.
Input
* Line 1: Two integers, N and P
* Lines 2..N: N-1 lines, each with two integers I and J. Node I is node J‘s parent in the tree of roads.
Output
A single line containing the integer that is the minimum number of roads that need to be destroyed for a subtree of P nodes to be isolated.
Sample Input
11 6 1 2 1 3 1 4 1 5 2 6 2 7 2 8 4 9 4 10 4 11
Sample Output
2
Hint
[A subtree with nodes (1, 2, 3, 6, 7, 8) will become isolated if roads 1-4 and 1-5 are destroyed.]
题意：有N个点和N-1条边构成的树，问你最少删去几条边使得新树中节点数为P。
思路：用dp[i][j]表示以i节点为根的树 中选中j个节点 最少需要删去的边数。
我们利用树形dp的思想，考虑u节点的子节点v。
一：直接去掉<u,v>边即去掉以v为根的子树，显然有dp[u][j] + 1；
二：保留边<u,v>，那么有dp[u][j-k] + dp[v][k],(1<= k <= j)。对于该情况我们要求出(1<=k<=j)范围的最小值再和第一种情况比较。
这样得到状态转移方程
dp[u][v] = min ( min(dp[u][j-k]+dp[v][k] ) , dp[u][v] + 1).
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define MAXN 200
#define INF 0x3f3f
using namespace std;
vector<int> G[MAXN];
int dp[MAXN][MAXN];//dp[i][j]存储以i节点为根的树 选j个点 最少删的边数
int pre[MAXN];
int N, P;
void init()
{
for(int i = 1; i <= N; i++)
G[i].clear(), pre[i] = i;
}
void getMap()
{
int a, b;
for(int i = 1; i < N; i++)
{
scanf("%d%d", &a, &b);
G[a].push_back(b);
pre[b] = a;
}
}
int num[MAXN];
void DFS(int u)
{
dp[u][1] = 0;//初始 自己不删边
for(int i = 0; i < G[u].size(); i++)
{
int v = G[u][i];
DFS(v);
for(int j = P; j >= 0; j--)
{
int t = dp[u][j] + 1;//直接删掉 与 子节点v 相连的边
for(int k = 0; k <= j; k++)
t = min(t, dp[u][j-k] + dp[v][k]);
dp[u][j] = t;
}
}
}
void solve()
{
int root;
for(int i = 1; i <= N; i++)
{
if(pre[i] == i)
{
root = i;
break;
}
}
memset(dp, INF, sizeof(dp));
DFS(root);
int ans = INF;
for(int i = 1; i <= N; i++)
{
if(i == root)//删边后 原根节点 依旧在
ans = min(ans, dp[i][P]);
else//删边后 原根节点已经没了 加上去掉根的一条边
ans = min(ans, dp[i][P] + 1);
}
printf("%d\n", ans);
}
int main()
{
while(scanf("%d%d", &N, &P) != EOF)
{
init();
getMap();
solve();
}
return 0;
}
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