POJ 1947 Rebuilding Roads 树形DP
2011-09-24 17:10
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题意:给出一棵树,至少切几下,能够得到一个包含p个节点的子树。
题解:假设节点u包含了cnt[u]个子节点:1,2,3,····cnt[u],每个子节点类似于一个背包,可以加或者不加。 当所有子节点都不加的时候,仅仅包含根节点一个,即dp[u][1]。那么我们需要对dp[u][1]进行一下初始化。即所有节点都不加的时候,只要切0次就得到根节点,即dp[u][1]=0。
贴一下错误的代码:
#include <iostream>
using namespace std;
#define N 155
#define INF 10000
#define min(a,b) (a<b?a:b)
int dp
, next
;
int cnt
, pre
;
int n, p;
void dfs ( int u )
{
if ( cnt[u] == 0 ) /* WR在这里·· */
{
dp[u][1] = 0;
return;
}
int i, j, k, v, temp;
for ( i = 1; i <= cnt[u]; i++ )
{
v = next[u][i];
dfs ( v );
for ( j = p; j >= 0; j-- )
{
temp = dp[u][j] + 1;
for ( k = 0; k <= j; k++ )
temp = min (dp[u][k]+dp[v][j-k], temp);
dp[u][j] = temp;
}
}
}
int main()
{
int i, j, x, y, ans;
while ( scanf("%d %d", &n, &p) != EOF )
{
memset(next,0,sizeof(next));
memset(cnt,0,sizeof(cnt));
memset(pre,0,sizeof(pre));
for ( i = 1; i < n; i++ )
{
scanf("%d %d",&x,&y);
next[x][++cnt[x]] = y;
pre[y] = 1;
}
for ( i = 0; i <= n; i++ )
for ( j = 0; j <= n; j++ )
dp[i][j] = INF;
for ( i = 1; i <= n; i++ )
if ( ! pre[i] ) break;
dfs ( i );
ans = dp[i][p];
for ( i = 2; i <= n; i++ )
if ( dp[i][p] < ans )
ans = dp[i][p] + 1;
printf("%d\n",ans);
}
return 0;
}
题解:假设节点u包含了cnt[u]个子节点:1,2,3,····cnt[u],每个子节点类似于一个背包,可以加或者不加。 当所有子节点都不加的时候,仅仅包含根节点一个,即dp[u][1]。那么我们需要对dp[u][1]进行一下初始化。即所有节点都不加的时候,只要切0次就得到根节点,即dp[u][1]=0。
#include <iostream> using namespace std; #define N 155 #define INF 10000 #define min(a,b) (a<b?a:b) int dp , next ;/* dp[i][j]表示以i为根节点,节点总数为j的子树,需要切的次数。 next[i][j]表示节点i的第j个子节点 */ int cnt , pre ; /* cnt[i]记录节点i的子节点数量。 pre[i]记录i的双亲节点 */ int n, p; void dfs ( int u ) { dp[u][1] = 0; int i, j, k, v, temp; for ( i = 1; i <= cnt[u]; i++ ) { v = next[u][i]; dfs ( v ); for ( j = p; j >= 0; j-- ) /* 类似于背包问题,不能 0 -> p */ { temp = dp[u][j] + 1; for ( k = 0; k <= j; k++ ) temp = min (dp[u][k]+dp[v][j-k], temp); dp[u][j] = temp; } } } int main() { int i, j, x, y, ans; while ( scanf("%d %d", &n, &p) != EOF ) { memset(next,0,sizeof(next)); memset(cnt,0,sizeof(cnt)); memset(pre,0,sizeof(pre)); for ( i = 1; i < n; i++ ) { scanf("%d %d",&x,&y); next[x][++cnt[x]] = y; pre[y] = 1; } for ( i = 0; i <= n; i++ ) for ( j = 0; j <= n; j++ ) dp[i][j] = INF; for ( i = 1; i <= n; i++ ) if ( ! pre[i] ) break;/* 双亲节点不存在的便是根节点 */ dfs ( i ); /*切完后的每一个节点都没有断绝与根节点的关系,那么需要在i与他的双亲节点之间再切一次(根节点除外)*/ ans = dp[i][p]; for ( i = 2; i <= n; i++ ) if ( dp[i][p] < ans ) ans = dp[i][p] + 1; printf("%d\n",ans); } return 0; }
贴一下错误的代码:
#include <iostream>
using namespace std;
#define N 155
#define INF 10000
#define min(a,b) (a<b?a:b)
int dp
, next
;
int cnt
, pre
;
int n, p;
void dfs ( int u )
{
if ( cnt[u] == 0 ) /* WR在这里·· */
{
dp[u][1] = 0;
return;
}
int i, j, k, v, temp;
for ( i = 1; i <= cnt[u]; i++ )
{
v = next[u][i];
dfs ( v );
for ( j = p; j >= 0; j-- )
{
temp = dp[u][j] + 1;
for ( k = 0; k <= j; k++ )
temp = min (dp[u][k]+dp[v][j-k], temp);
dp[u][j] = temp;
}
}
}
int main()
{
int i, j, x, y, ans;
while ( scanf("%d %d", &n, &p) != EOF )
{
memset(next,0,sizeof(next));
memset(cnt,0,sizeof(cnt));
memset(pre,0,sizeof(pre));
for ( i = 1; i < n; i++ )
{
scanf("%d %d",&x,&y);
next[x][++cnt[x]] = y;
pre[y] = 1;
}
for ( i = 0; i <= n; i++ )
for ( j = 0; j <= n; j++ )
dp[i][j] = INF;
for ( i = 1; i <= n; i++ )
if ( ! pre[i] ) break;
dfs ( i );
ans = dp[i][p];
for ( i = 2; i <= n; i++ )
if ( dp[i][p] < ans )
ans = dp[i][p] + 1;
printf("%d\n",ans);
}
return 0;
}
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