[树形DP] SRM 598 Div1 Hard TPS

最新推荐文章于 2019-04-12 14:10:19 发布

原创最新推荐文章于 2019-04-12 14:10:19 发布 · 394 阅读

1 ·

CC 4.0 BY-SA版权

动态规划同时被 2 个专栏收录

62 篇文章

订阅专栏

树

22 篇文章

订阅专栏

本文介绍了一种利用树形动态规划解决信标放置问题的方法，通过枚举树的根节点并放置信标来区分不同深度的节点，确保能够区分所有节点，同时最小化信标的使用数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

$Solution$

可以枚举根，在根放一个信标，就可以区分出深度不同的点。

若在节点 $u$ 放信标，则可以区分出点 $v$ 是否在 $u$ 内。可以考虑从 $dis_{u,v}=dep_u+dep_v-2dep_{lca(u,v)}$ 求出 $dep_{lca(u,v)}$ ，再与 $dep_u$ 比较即可。
若节点 $u$ 的两个子树 $c_1,c_2$ 内部都没有信标，显然无法区分两子树中深度相同的点。
若一个节点的儿子集合大小为 $|C|$ ，至少需要在 $|C|-1$ 个里面放信标。

枚举 $root$ ，树形DP即可。

// BEGIN CUT HERE

// END CUT HERE
#line 5 "TPS.cpp"
#include <bits/stdc++.h>
using namespace std;

const int N = 100;
const int INF = 1 << 30;

int mp[N][N];
int n, ans;
int f[N];

class TPS {

public:
    inline void dp(int u, int fa) {
        f[u] = 0;
        for (int to = 1; to <= n; to++)
            if (mp[u][to] && to != fa) {
                dp(to, u); f[u] += max(f[to], 1);
            }
        for (int to = 1; to <= n; to++)
            if (mp[u][to] && to != fa)
                if (f[to] == 0) {
                    --f[u]; break;
                }
    }
    int minimalBeacons(vector <string> linked) {
        n = linked.size();
        if (n == 1) return 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                mp[i + 1][j + 1] = (linked[i][j] == 'Y');
        ans = INF;
        for (int i = 1; i <= n; i++) {
            dp(i, 0); ans = min(f[i] + 1, ans);
        }
        return ans;
    }

// BEGIN CUT HERE
    public:
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) test_case_5(); }
    private:
    template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
    void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
    void test_case_0() { string Arr0[] = {"NYNN",
 "YNYN",
 "NYNY",
 "NNYN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; verify_case(0, Arg1, minimalBeacons(Arg0)); }
    void test_case_1() { string Arr0[] = {"NYYY",
 "YNNN",
 "YNNN",
 "YNNN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 2; verify_case(1, Arg1, minimalBeacons(Arg0)); }
    void test_case_2() { string Arr0[] = {"NNYNNNNNNN",
 "NNNNNYNNNN",
 "YNNYNNYNNN",
 "NNYNYNNYNN",
 "NNNYNYNNYN",
 "NYNNYNNNNY",
 "NNYNNNNNNN",
 "NNNYNNNNNN",
 "NNNNYNNNNN",
 "NNNNNYNNNN"}
; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 2; verify_case(2, Arg1, minimalBeacons(Arg0)); }
    void test_case_3() { string Arr0[] = {"NYNYNNYNN",
 "YNYNYNNYN",
 "NYNNNYNNY",
 "YNNNNNNNN",
 "NYNNNNNNN",
 "NNYNNNNNN",
 "YNNNNNNNN",
 "NYNNNNNNN",
 "NNYNNNNNN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 3; verify_case(3, Arg1, minimalBeacons(Arg0)); }
    void test_case_4() { string Arr0[] = {"NYYYYYYYYY",
 "YNNNNNNNNN",
 "YNNNNNNNNN",
 "YNNNNNNNNN",
 "YNNNNNNNNN",
 "YNNNNNNNNN",
 "YNNNNNNNNN",
 "YNNNNNNNNN",
 "YNNNNNNNNN",
 "YNNNNNNNNN"}
; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 8; verify_case(4, Arg1, minimalBeacons(Arg0)); }
    void test_case_5() { string Arr0[] = {"N"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 0; verify_case(5, Arg1, minimalBeacons(Arg0)); }

// END CUT HERE


};

// BEGIN CUT HERE
int main(void) {
    TPS ___test;
    ___test.run_test(-1);
    system("pause");
}
// END CUT HERE