UVa 1220 Party at Hali-bula (tree DP, max standalone set)

Test instructions: The company has n people to form a tree structure, in addition to the boss have only a direct boss, ask to choose as many as possible, but not at the same time select a person and his direct supervisor, ask the maximum number of people can choose, and is not the only solution.

Analysis: This question is almost the biggest independent set problem of the tree, but one more judgment uniqueness. Use two arrays, one to record the number of people, one to judge uniqueness.

D[u][0], in a subtree that is the root of U, the maximum number of people can be obtained without selecting u points, then d[u][1] is the maximum number of points that can be selected.

F[u][0] is similar, in a subtree that is rooted in u, whether the u point is unique or not, then f[u][1] is the choice of whether the U point is unique.

For d[u][1] calculations, because you chose U, then the child nodes of U are not selectable, so it is d[u][1] = SUM (d[v][0], V is a child node), when F[v][0] is not unique, f[u][1] is not unique.

For d[u][0] calculation, because no U is selected, then its sub-nodes can be selected or not selected, that is, the largest choice, that is, d[u][0] = SUM (max (d[v][0), d[v][1]), then the uniqueness of how to judge it? Almost like the top,

One more, if d[v][0] = = D[v][1], then this is not the only, the other and the same as above. The rest is simple, with DFS.

The code is as follows:

#include <iostream> #include <cstdio> #include <map> #include <cstring> #include <vector>  using namespace Std;const int maxn = + 5;vector<int> g[maxn];bool f[maxn][2];int d[maxn][2], N, cnt;//d[u][0] not selected    , d[u][0] Select map<string, int> id;void init () {//Initialize for (int i = 1; I <= n; ++i) g[i].clear ();    Memset (f, false, sizeof (f));    memset (d, 0, sizeof (d));   CNT = 0; Id.clear ();}    int GetID (const string &s) {//Get ID if (Id.count (s)) return id[s]; return id[s] = ++cnt;} void Dfs (int u) {if (!        G[u].size ()) {//bottom d[u][0] = 0;        D[U][1] = 1;    return;        } for (int i = 0; i < g[u].size (); ++i) {int son = g[u][i];        DFS (son); D[U][1] + = d[son][0];//d[u][1] calculation if (f[son][0]) f[u][1] = true;//judgment uniqueness if (D[son][0] > D[son][1]) {//d[u]            [0] Calculation d[u][0] + = d[son][0];        if (f[son][0]) f[u][0] = true; } else if (d[son][0] = = D[son][1]) {//equal, that is not unique d[u][0] + = d[son][0];        F[u][0] = true;            } else {d[u][0] + = d[son][1];        if (f[son][1]) f[u][0] = true;        }} ++d[u][1];//don't forget to add 1}int main () {while (scanf ("%d", &n) = = 1 && N) {init ();        string s1, S2;  CIN >> S1;        GetID (S1);            for (int i = 0; i < n-1; ++i) {cin >> s1 >> S2;        G[getid (S2)].push_back (GetID (S1));        } dfs (1); if (d[1][0] = = d[1][1]) printf ("%d no\n", d[1][0]);//judge who is in the number greater else if (d[1][0] > D[1][1]) printf ("%d%s\n", d[1][ 0], f[1][0]?        "No": "Yes"); else printf ("%d%s\n", d[1][1], f[1][1]?    "No": "Yes"); } return 0;}

UVa 1220 Party at Hali-bula (tree DP, max standalone set)