我是菜鸡，我是菜鸡，我是菜鸡。。。。重要的事说三遍
算是第一次做树形dp的题吧，不太难。。
园林构成一棵树，root为1，Hi从root出发，有k个园林必须玩，每个园林游玩后会得到权值w[i]，最多玩M个园林。
经过的园林必须玩，问可得到的最大权值和。
题目链接：http://hihocoder.com/problemset/problem/1104
经典的树形dp，dp[i][j]，表示以i为根的子树选j个结点可得的最大权值和。
这样dp[root][num] = g[son_num][num-1] + w[root];
g[son_num][num-1]类似于对root的儿子结点的子树做背包，g[i][j] = max( g[i-1][j-p] + dp[child_id][p] )；
题解不错：http://hihocoder.com/discuss/question/2743
对于必须玩的k个点，他的处理方法我没有看懂。。。。discussion里的不错。。。游玩结点i，那么其祖先结点fa[i]必要也游玩，将这些点设为must，
将must构成的子树缩成一个点，rebuild a new tree，就成了上面说的最普通的树形dp了。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 const int maxn = 100 + 10; 5 int n, k, m; 6 7 vector<int> G[maxn], G1[maxn]; 8 int fa[maxn], w[maxn], must[maxn], vis[maxn]; 9 int f[maxn][maxn]; 10 11 void init(){ 12 for( int i = 0; i < maxn; ++i ){ 13 G[i].clear(); 14 G1[i].clear(); 15 } 16 memset( fa, -1, sizeof(fa) ); 17 memset( f, -1, sizeof(f) ); 18 memset( must, 0, sizeof(must) ); 19 } 20 21 void dfs( int u, int pa ){ 22 fa[u] = pa; 23 for( int i = 0; i < G[u].size(); ++i ){ 24 int v = G[u][i]; 25 if( v == pa ) 26 continue; 27 dfs( v, u ); 28 } 29 } 30 31 void dfs1( int u, int pa ){ 32 if( pa != -1 ){ 33 if( vis[pa] && !vis[u] ) 34 G1[1].push_back(u); 35 else if( !vis[pa] && !vis[u] ) 36 G1[pa].push_back(u); 37 } 38 39 for( int i = 0; i < G[u].size(); ++i ){ 40 int v = G[u][i]; 41 if( v == pa ) 42 continue; 43 dfs1( v, u ); 44 } 45 } 46 47 int dp( int root, int num ){ 48 if( num == 0 ) 49 return 0; 50 if( f[root][num] != -1 ) 51 return f[root][num]; 52 53 int g[maxn][maxn], son_num = G1[root].size(); 54 memset( g, 0, sizeof(g) ); 55 for( int i = 1; i <= son_num; ++i ){ 56 int child = G1[root][i-1]; 57 for( int j = 0; j < num; ++j ){ 58 for( int p = 0; p <= j; ++p ){ 59 g[i][j] = max( g[i][j], g[i-1][j-p] + dp( child, p ) ); 60 } 61 } 62 } 63 64 f[root][num] = g[son_num][num-1] + w[root]; 65 return f[root][num]; 66 } 67 68 void print( int num ){ 69 for( int i = 0; i < G1[num].size(); ++i ) 70 cout << G1[num][i] << " "; 71 cout << endl; 72 } 73 74 void solve(){ 75 dfs(1, -1); 76 77 //must节点压缩 78 int ans = 0, cnt = 0; 79 memset( vis, 0, sizeof(vis) ); 80 for( int i = 1; i <= n; ++i ){ 81 if(must[i]){ 82 int u = i; 83 while(u != -1 && !vis[u]){ 84 vis[u] = 1; 85 cnt++, ans += w[u]; 86 u = fa[u]; 87 } 88 } 89 } 90 //cout << "ans: " << ans << endl; 91 92 if( cnt > m ){ 93 printf("-1 "); 94 return; 95 } 96 97 //rebuild tree 98 dfs1( 1, -1 ); 99 100 w[1] = ans; 101 //cout << "m-cnt: " << m - cnt << endl; 102 ans = dp( 1, m-cnt+1 ); 103 //cout << "f: " << f[8][3] << endl; 104 105 printf("%d ", ans); 106 } 107 108 int main(){ 109 //freopen("1.in", "r", stdin); 110 init(); 111 scanf("%d%d%d", &n, &k, &m); 112 for( int i = 1; i <= n; ++i ){ 113 scanf("%d", &w[i]); 114 } 115 116 int t; 117 for( int i = 1; i <= k; ++i ){ 118 scanf("%d", &t); 119 must[t] = 1; 120 } 121 122 int a, b; 123 for( int i = 1; i <= n-1; ++i ){ 124 scanf("%d%d", &a, &b); 125 G[a].push_back(b), G[b].push_back(a); 126 } 127 128 solve(); 129 130 return 0; 131 }
当然，也可以不需要g数组，直接当一维背包来做
dp[root][x] = max( dp[root][x], dp[root][x-y] + dp[child_id][y] );
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 const int maxn = 100 + 10; 5 int n, k, m; 6 7 vector<int> G[maxn], G1[maxn]; 8 int fa[maxn], w[maxn], must[maxn], vis[maxn]; 9 int f[maxn][maxn]; 10 11 void init(){ 12 for( int i = 0; i < maxn; ++i ){ 13 G[i].clear(); 14 G1[i].clear(); 15 } 16 memset( fa, -1, sizeof(fa) ); 17 memset( f, -1, sizeof(f) ); 18 memset( must, 0, sizeof(must) ); 19 } 20 21 void dfs( int u, int pa ){ 22 fa[u] = pa; 23 for( int i = 0; i < G[u].size(); ++i ){ 24 int v = G[u][i]; 25 if( v == pa ) 26 continue; 27 dfs( v, u ); 28 } 29 } 30 31 void dfs1( int u, int pa ){ 32 if( pa != -1 ){ 33 if( vis[pa] && !vis[u] ) 34 G1[1].push_back(u); 35 else if( !vis[pa] && !vis[u] ) 36 G1[pa].push_back(u); 37 } 38 39 for( int i = 0; i < G[u].size(); ++i ){ 40 int v = G[u][i]; 41 if( v == pa ) 42 continue; 43 dfs1( v, u ); 44 } 45 } 46 47 void dp(int root, int pa){ 48 f[root][1] = w[root]; 49 for( int i = 0; i < G1[root].size(); ++i ){ 50 int v = G1[root][i]; 51 if( v == pa ) 52 continue; 53 dp( v, root ); 54 for( int x = m; x >= 1; --x ){ 55 for( int y = 0; y < x; ++y ){ 56 f[root][x] = max( f[root][x], f[root][x-y] + f[v][y] ); 57 } 58 } 59 } 60 } 61 62 void solve(){ 63 dfs(1, -1); 64 65 //must½ÚµãѹËõ 66 int ans = 0, cnt = 0; 67 memset( vis, 0, sizeof(vis) ); 68 for( int i = 1; i <= n; ++i ){ 69 if(must[i]){ 70 int u = i; 71 while(u != -1 && !vis[u]){ 72 vis[u] = 1; 73 cnt++, ans += w[u]; 74 u = fa[u]; 75 } 76 } 77 } 78 //cout << "ans: " << ans << endl; 79 80 if( cnt > m ){ 81 printf("-1 "); 82 return; 83 } 84 85 //rebuild tree 86 dfs1( 1, -1 ); 87 88 w[1] = ans; 89 dp(1, -1); 90 printf("%d ", f[1][m-cnt+1]); 91 } 92 93 int main(){ 94 //freopen("1.in", "r", stdin); 95 init(); 96 scanf("%d%d%d", &n, &k, &m); 97 for( int i = 1; i <= n; ++i ){ 98 scanf("%d", &w[i]); 99 } 100 101 int t; 102 for( int i = 1; i <= k; ++i ){ 103 scanf("%d", &t); 104 must[t] = 1; 105 } 106 107 int a, b; 108 for( int i = 1; i <= n-1; ++i ){ 109 scanf("%d%d", &a, &b); 110 G[a].push_back(b), G[b].push_back(a); 111 } 112 113 solve(); 114 115 return 0; 116 }