HDU 2412 party at Hali-Bula (tree DP, map)

N people form a relationship tree. Each node represents a person. The parent node of the node represents the only boss of the person. Only the root node has no boss. It is required to select a portion of the N people so that no direct upper-lower-level relationship exists between any two people. How many people can be selected at most? Determine whether the optimal solution is unique. If yes, yes. Otherwise, no.

Question: DP [I] [0] indicates the maximum number of people that can be selected from the subtree with the root node when I is not selected; DP [I] [1] indicates that when I is selected, the maximum number of people that can be selected from the subtree with it as the root. Sole [I] [0] indicates whether the optimal solution is unique when I is not selected; sole [I] [1] indicates whether the optimal solution is unique when I is selected.

PS: There are detailed instructions on this PPT http://wenku.baidu.com/view/84164e1a227916888486d7d6.html? From = rec & Pos = 4 & Weight = 1

#include<map>#include<cstring>#include<string>#include<algorithm>#include<iostream>using namespace std;map<string,int> name;map<string,int>::iterator it;int dp[500][2];bool sole[500][2], visit[500];struct Edge { int v, next; } edge[500];int head[500], E, n;void add_edge(int u, int v){    edge[E].v = v;    edge[E].next = head[u];    head[u] = E++;}void DFS(int u, int father){    if(visit[u]) return;    if(head[u] == -1)    {        dp[u][0] = 0;        dp[u][1] = 1;        sole[u][0] = sole[u][1] = true;        visit[u] = true;        return;    }    visit[u] = true;    dp[u][0] = 0; dp[u][1] = 1;    sole[u][0] = sole[u][1] = true;    for(int i = head[u]; i != -1; i = edge[i].next)    {        int v = edge[i].v;        if(v == father) continue;        if(!visit[v]) DFS(v, u);        dp[u][0] += max(dp[v][0], dp[v][1]);        dp[u][1] += dp[v][0];        if(dp[v][0] > dp[v][1] && !sole[v][0]) sole[u][0] = false;        if(dp[v][0] < dp[v][1] && !sole[v][1]) sole[u][0] = false;        if(dp[v][0] == dp[v][1]) sole[u][0] = false;        if(sole[v][0] == false) sole[u][1] = false;    }}int main(){    string employee, boss;    while(cin >> n)    {        if(n == 0) break;        int i, c, u, v;        c = E = 0;        memset(head, -1, sizeof(head));        memset(visit, 0, sizeof(visit));        name.clear();        cin >> boss;        name.insert(pair<string, int>(boss, 0));        for(i = 1; i < n; i++)        {            cin >> employee >> boss;            it = name.find(employee);            if(it ==  name.end())            {                name.insert(pair<string, int>(employee, ++c));                u = c;            }            else u = it->second;            it = name.find(boss);            if(it ==  name.end())            {                name.insert(pair<string, int>(boss, ++c));                v = c;            }            else v = it->second;            add_edge(u, v);            add_edge(v, u);        }        DFS(0, -1);        if(dp[0][0] > dp[0][1])            cout << dp[0][0] << (sole[0][0] ? " Yes" : " No") << endl;        else if(dp[0][0] < dp[0][1])            cout << dp[0][1] << (sole[0][1] ? " Yes" : " No") << endl;        else            cout << dp[0][0] << " No" << endl;    }}