树形dp题，状态转移方程应该很好推，但一定要细心。
http://poj.org/problem?id=3342
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <string> 5 #include <map> 6 using namespace std; 7 const int maxn = 200 + 10; 8 char buf[maxn]; 9 struct Edge{ 10 int to, next; 11 }edge[maxn << 1]; 12 int head[maxn], N; 13 map<string, int> mapi; 14 int n, k; 15 int dp[maxn][2]; 16 int o[maxn][2]; 17 int get_str(char *dest){ 18 int cnt = 0; 19 char ch; 20 do{ 21 ch = getchar(); 22 }while(ch != EOF && (ch == ' ' || ch == ' ')); 23 if(ch == EOF) return 0; 24 dest[cnt++] = ch; 25 while((ch = getchar()) != ' ' && ch != ' ' && ch != EOF) dest[cnt++] = ch; 26 dest[cnt] = ' '; 27 return cnt; 28 } 29 30 31 void addEdge(int u, int v){ 32 edge[N].next = head[u]; 33 edge[N].to = v; 34 head[u] = N++; 35 } 36 37 void dfs(int u, int fa){ 38 dp[u][1] = 1; 39 int tem1 = 0, tem2 = 0; 40 for(int i = head[u]; i + 1; i = edge[i].next){ 41 int v = edge[i].to; 42 if(v == fa) continue; 43 dfs(v, u); 44 dp[u][1] += dp[v][0]; 45 o[u][1] |= o[v][0]; 46 int d = dp[v][1] > dp[v][0] ? 1 : (dp[v][1] == dp[v][0] ? -1 : 0); 47 if(d == -1){ 48 dp[u][0] += dp[v][0]; 49 o[u][0] = 1; 50 }else{ 51 dp[u][0] += dp[v][d]; 52 o[u][0] |= o[v][d]; 53 } 54 } 55 } 56 57 int main(){ 58 //freopen("in.txt", "r", stdin); 59 while(~scanf("%d", &n) && n){ 60 k = 0; 61 int base = 0; 62 get_str(buf); 63 mapi.clear(); 64 mapi[string(buf)] = ++base; 65 N = 0; 66 memset(head, -1, sizeof head); 67 for(int i = 1, x, y; i < n; i++){ 68 get_str(buf); 69 if(mapi.find(string(buf)) == mapi.end()) mapi[string(buf)] = ++base, x = base; 70 else x = mapi[string(buf)]; 71 get_str(buf); 72 if(mapi.find(string(buf)) == mapi.end()) mapi[string(buf)] = ++base, y = base; 73 else y = mapi[string(buf)]; 74 addEdge(x, y); 75 addEdge(y, x); 76 } 77 memset(dp, 0, sizeof dp); 78 memset(o, 0, sizeof o); 79 dfs(1, 0); 80 int ans = -1; 81 bool ok = 0; 82 for(int i = 1; i <= n; i++) for(int j = 0; j <= 1; j++){ 83 if(dp[i][j] > ans) ans = dp[i][j], ok = o[i][j]; 84 else if(dp[i][j] == ans) ok = 1; 85 } 86 printf("%d %s ", ans, ok ? "No" : "Yes"); 87 } 88 return 0; 89 }