Rebuilding Roads
Time Limit:1000MSMemory Limit:30000K
Total Submissions:9496Accepted: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: 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 61 21 31 41 52 62 72 84 94 104 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.]
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 0x3f3f3f3fstruct 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] == -1 ){c[u][ sum[u] ] = 0 ;return ;}int i , j , k , v , temp ;for(i = head[u] ; i != -1 ; i = edge[i].next) {v = edge[i].v ;dfs(v) ;sum[u] += sum[v] ;}c[u][ sum[u] ] = 0 ;for(i = head[u] ; i != -1 ; 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,-1,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 < n-1 ; 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\n", min1) ;return 0 ;}
不会因为忧伤而风情万种。
