题意:给定一棵树,求大小为k的一个子树的最大权值。
析:dp[i][j] 表示以 i 为根大小为 j 时最大权值。dp[i][j] = max{dp[i][j-k] + dp[son][k]},状态方程。
有一个要注意,因为要选的是一棵子树,所以以哪个点为根都行,也就是说,对于任意子树都能找一个合适的根,使得不会出现父结点与子结点冲突。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-6; const int maxn = 100 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r > 0 && r <= n && c > 0 && c <= m; } int dp[maxn][maxn]; int val[maxn]; struct Edge{ int to, next; }; int head[maxn], cnt; Edge edge[maxn<<1]; void add_edge(int u, int v){ edge[cnt].to = v; edge[cnt].next = head[u]; head[u] = cnt++; } int ans; void dfs(int u, int fa){ dp[u][1] = val[u]; for(int i = head[u]; ~i; i = edge[i].next){ int v = edge[i].to; if(v == fa) continue; dfs(v, u); for(int j = m; j > 1; --j) for(int k = 1; k < j; ++k) dp[u][j] = max(dp[u][j], dp[u][j-k] + dp[v][k]); } ans = max(ans, dp[u][m]); } int main(){ while(scanf("%d %d", &n, &m) == 2){ for(int i = 0; i < n; ++i) scanf("%d", val+i); memset(head, -1, sizeof head); cnt = 0; for(int i = 1; i < n; ++i){ int u, v; scanf("%d %d", &u, &v); add_edge(u, v); add_edge(v, u); } memset(dp, 0, sizeof dp); ans = 0; dfs(0, -1); printf("%d\n", ans); } return 0; }
时间: 2024-10-10 15:02:23