【TOJ 3522.】Tree cutting【树形DP】
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题意:给出一棵树,节点有价值和重量,选取一个联通分量使得总价值/总重量最大,并且这个联通分量的点数大于等于k。
思路:树形DP,dpv[i][j], dpw[i][j]分别表示以i为根的子树选取了j个点时最佳值的价值和重量(包括i点),然后一遍dfs就可以把所有的dp求出来,再在里面求出满足条件的最优解即可。
#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;#define N 102vector<int>V[N];int n, k, val[N], we[N];int dpv[N][N], dpw[N][N], sun[N];double ans;void dfs(int u, int f) { int i, j, h, v; sun[u] = 1; dpv[u][1] = val[u], dpw[u][1] = we[u]; for (i = 0;i < V[u].size();i++) { v = V[u][i]; if (v == f) continue; dfs(v, u); sun[u] += sun[v]; for (j = sun[u];j >= 1;j--) { for (h = 1;h <= sun[v] && h < j;h++) { if (dpv[u][j-h] == -1) continue; if (dpv[v][h] == -1) continue; if (dpv[u][j] == -1) { dpv[u][j] = dpv[u][j-h]+dpv[v][h], dpw[u][j] = dpw[u][j-h]+dpw[v][h]; }else { int tv = dpv[u][j-h]+dpv[v][h], tw = dpw[u][j-h]+dpw[v][h]; if ((long long)tw*dpv[u][j] < (long long)tv*dpw[u][j]) dpv[u][j] = tv, dpw[u][j] = tw; } } } } for (i = k;i <= sun[u];i++) { if (dpv[u][i] == -1) continue; double tm = 1.0*dpv[u][i]/dpw[u][i]; if (ans < tm) ans = tm; }}int main() { int T, i, u, v; scanf("%d", &T); while (T--) { scanf("%d%d", &n, &k); for (i = 1;i <= n;i++) V[i].clear(); for (i = 1;i <= n;i++) scanf("%d", &val[i]); for (i = 1;i <= n;i++) scanf("%d", &we[i]); for (i = 1;i < n;i++) { scanf("%d%d", &u, &v); V[u].push_back(v), V[v].push_back(u); } ans = 0.0; memset(sun, 0, sizeof(sun)); memset(dpv, -1, sizeof(dpv)); memset(dpw, -1, sizeof(dpw)); dfs(1, -1); printf("%.2lf\n", ans); }}
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