poj1947--Rebuilding Roads(樹狀dp)
來源：程序員人生 發布時間：2015-03-25 11:56:46 閱讀次數：3444次
Rebuilding Roads
| Time Limit: 1000MS |
|
Memory Limit: 30000K |
| Total Submissions: 9496 |
|
Accepted: 4316 |
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: N⑴ 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 1⑷ and 1⑸ are destroyed.]
Source
USACO 2002 February
給出n個節點的樹，給出值m。問最少刪除幾條邊可以得到節點個數為m的子樹。
樹狀dp，統計出以節點i為根的子樹得到節點個數為j的子樹最少刪除的邊數。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define INF 0x3f3f3f3f
struct node {
int v , next ;
}edge[160] ;
int head[160] , cnt ;
int c[160][160] , sum[160];
void add(int u,int v) {
edge[cnt].v = v ;
edge[cnt].next = head[u] ;
head[u] = cnt++ ;
}
void dfs(int u)
{
sum[u] = 1 ;
if( head[u] == ⑴ )
{
c[u][ sum[u] ] = 0 ;
return ;
}
int i , j , k , v , temp ;
for(i = head[u] ; i != ⑴ ; i = edge[i].next) {
v = edge[i].v ;
dfs(v) ;
sum[u] += sum[v] ;
}
c[u][ sum[u] ] = 0 ;
for(i = head[u] ; i != ⑴ ; i = edge[i].next) {
v = edge[i].v ;
c[v][0] = 1 ;
for(j = 0 ; j <= sum[u] ; j++) {
for(k = 0 ; k <= sum[v] ; k++) {
temp = sum[v] - k ;
if( j >= temp )
c[u][ j-temp ] = min( c[u][j-temp],c[u][j]+c[v][k] ) ;
}
}
c[v][0] = INF ;
}
return ;
}
int main() {
int n , p , i , u , v ;
memset(head,⑴,sizeof(head)) ;
memset(c,INF,sizeof(c)) ;
memset(sum,0,sizeof(sum)) ;
cnt = 0 ;
scanf("%d %d", &n, &p) ;
add(0,1) ;
for(i = 0 ; i < n⑴ ; i++) {
scanf("%d %d", &u, &v) ;
add(u,v) ;
}
dfs(0) ;
int min1 = c[1][p] ;
for(i = 2 ; i <= n ; i++) {
min1 = min(min1,c[i][p]+1) ;
}
printf("%d
", min1) ;
return 0 ;
}
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