题目:给你一棵树。找到最小的顶点集合,使得全部的边至少有一个顶点在这个集合中。
分析:树形dp,图论,最小顶点覆盖。
方案1:树形dp。分别记录每一个节点取和不取的最优解f(k。0)与f(k,1);
每一个节点的状态取决于子树,子树的根都不选,则他必选;否则取最小;
f(k。0)= sum(f(i,1))。
f(k。1)= sum(min(f(i,0)。f(i。1))){ 当中 i 是k的子树根节点 };
方案2:最小顶点覆盖 = N - 最大独立点 = 最大匹配。
说明:(2011-09-19 09:46)。
#include <stdio.h>
#include <stdlib.h>
#include <string.h> #define min(x,y) ((x)<(y)?(x):(y)) typedef struct node
{
int Count;
int Value;
int Next[ 10 ];
}node;
node Node[ 1510 ];
int Root; typedef struct answ
{
int sum0;
int sum1;
}Answ;
Answ Save; Answ dp( int Root )
{
Answ an;
an.sum0 = 0;
an.sum1 = 1;
if ( Node[ Root ].Count ) {
for ( int i = 0 ; i < Node[ Root ].Count ; ++ i ) {
Save = dp( Node[ Root ].Next[ i ] );
an.sum0 += Save.sum1;
an.sum1 += min( Save.sum0, Save.sum1 );
}
}
return an;
} int main()
{
int n,a,m,b;
while ( scanf("%d",&n) != EOF ) {
Root = -1;
memset( Node, 0, sizeof( Node ) );
for ( int i = 0 ; i < n ; ++ i ) {
scanf("%d:(%d)",&a,&m);
if ( Root == -1 ) Root = a;
Node[ a ].Count = m;
Node[ a ].Value = a;
for ( int j = 0 ; j < m ; ++ j ) {
scanf("%d",&b);
Node[ a ].Next[ j ] = b;
}
}
Answ answer = dp( Root );
printf("%d\n",min( answer.sum0, answer.sum1 ));
}
return 0;
}
图论解法:
#include <stdio.h>
#include <stdlib.h> typedef struct node
{
int Point;
node* Next;
}node;
node Node[ 3001 ];
node *Head[ 1501 ];
bool Used[ 1501 ];
int Result[ 1501 ]; bool find( int a, int n )
{
for ( node *P = Head[ a ] ; P ; P = P->Next )
if ( !Used[ P->Point ] ) {
Used[ P->Point ] = true;
if ( Result[ P->Point ] == -1 || find( Result[ P->Point ] , n ) ) {
Result[ P->Point ] = a;
return true;
}
}
return false;
} int argument( int n )
{
for ( int i = 0 ; i < n ; ++ i )
Result[ i ] = -1;
int Count = 0;
for ( int i = 0 ; i < n ; ++ i ) {
for ( int j = 0 ; j < n ; ++ j )
Used[ j ] = false;
if ( find( i, n ) )
++ Count;
}
return Count;
} int main()
{
int n,a,m,b;
while ( scanf("%d",&n) != EOF ) {
for ( int i = 0 ; i < n ; ++ i )
Head[ i ] = NULL;
int Count = 0;
for ( int i = 0 ; i < n ; ++ i ) {
scanf("%d:(%d)",&a,&m);
for ( int j = 0 ; j < m ; ++ j ) {
scanf("%d",&b);
Node[ Count ].Next = Head[ a ];
Node[ Count ].Point = b;
Head[ a ] = &Node[ Count ++ ];
Node[ Count ].Next = Head[ b ];
Node[ Count ].Point = a;
Head[ b ] = &Node[ Count ++ ];
}
}
printf("%d\n",argument( n )/2);
}
return 0;
}