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很简单的树形dp题目,转移方程是:
dp[u][0] += dp[v][1];
dp[u][1] += min( dp[v][0], dp[v][1] );
其中u是v的父亲节点。
1 #include <algorithm> 2 #include <iostream> 3 #include <cstring> 4 #include <cstdio> 5 using namespace std; 6 7 const int N = 1500; 8 bool flag[N]; 9 int head[N]; 10 int dp[N][2]; 11 int e, n; 12 13 void init() 14 { 15 e = 0; 16 memset( flag, true, sizeof(flag) ); 17 memset( head, -1, sizeof(head) ); 18 } 19 20 struct Edge 21 { 22 int v, next; 23 } edge[N]; 24 25 void addEdge( int u, int v ) 26 { 27 edge[e].v = v; 28 edge[e].next = head[u]; 29 head[u] = e++; 30 } 31 32 void dfs( int u ) 33 { 34 dp[u][0] = 0; 35 dp[u][1] = 1; 36 for ( int i = head[u]; i != -1; i = edge[i].next ) 37 { 38 int v = edge[i].v; 39 dfs(v); 40 dp[u][0] += dp[v][1]; 41 dp[u][1] += min( dp[v][0], dp[v][1] ); 42 } 43 } 44 45 int main () 46 { 47 while ( scanf("%d", &n) != EOF ) 48 { 49 init(); 50 for ( int i = 0; i < n; i++ ) 51 { 52 int u, v, m; 53 scanf("%d:(%d)", &u, &m); 54 while ( m-- ) 55 { 56 scanf("%d", &v); 57 addEdge( u, v ); 58 flag[v] = false; 59 } 60 } 61 int root; 62 for ( int i = 0; i < n; i++ ) 63 { 64 if ( flag[i] ) 65 { 66 root = i; 67 break; 68 } 69 } 70 dfs(root); 71 int ans = min( dp[root][0], dp[root][1] ); 72 printf("%d\n", ans); 73 } 74 return 0; 75 }
也可以用二分图来做。
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原文地址:http://www.cnblogs.com/huoxiayu/p/4648101.html