poj1947--Rebuilding Roads(树状dp)
发布时间：2020-12-13 20:17:25 所属栏目：PHP教程 来源：网络整理
导读：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. T
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Rebuilding Roads
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
给出n个节点的树，给出值m。问最少删除几条边可以得到节点个数为m的子树。 树状dp，统计出以节点i为根的子树得到节点个数为j的子树最少删除的边数。
#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,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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