UVA1220Party at Hali-Bula(树的最大独立集 + 唯一性判断)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=105116#problem/H
紫书P282
员工和直属老板只能选一个,最多选多少人
思路:d(u,0)表示以U为根的子树,不选u点,则子节点可选可不选,f(u,0)表示不选u的唯一性 (1表示唯一,0,表示不唯一)
d(u,1)以u为根的子树,选择u点,f(u,1)表示选择u的唯一性
转移方程:d(u,0) = sum{ max( d(v,0), d(v,1) ) },v是子节点,当d(v,0) == d(v,1) ,不唯一,或者选择的那个不唯一,则f(u,0)不唯一
d(u,1) = sum{ d(v,0) }, 当f(v,0)有一个不唯一,则不唯一
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cstdio> 5 #include <vector> 6 #include <map> 7 using namespace std; 8 const int MAX = 210; 9 vector <int> son[MAX]; 10 map<string, int> name; 11 int d[MAX][2],f[MAX][2]; 12 void DP(int u) 13 { 14 if(son[u].size() == 0) 15 { 16 d[u][1] = f[u][1] = 1; 17 d[u][0] = 0; 18 f[u][0] = 1; 19 return; 20 } 21 int c = (int) son[u].size(); 22 for(int i = 0; i < c; i++) 23 { 24 DP(son[u][i]); 25 } 26 int sum = 0,flag = 0; 27 for(int i = 0; i < c; i++) 28 { 29 sum += d[ son[u][i] ][0]; 30 if(f[ son[u][i] ][0] == 0) 31 flag = 1; 32 } 33 if(flag) 34 f[u][1] = 0; 35 else 36 f[u][1] = 1; 37 d[u][1] = max(d[u][1], sum + 1); 38 sum = 0,flag = 0; 39 for(int i = 0; i < c; i++) 40 { 41 if(d[ son[u][i] ][1] > d[ son[u][i] ][0]) 42 { 43 sum += d[ son[u][i] ][1] ; 44 if(f[ son[u][i] ][1] == 0) 45 flag = 1; 46 } 47 else if(d[ son[u][i] ][1] < d[ son[u][i] ][0]) 48 { 49 sum += d[ son[u][i] ][0]; 50 if(f[ son[u][i] ][0] == 0) 51 flag = 1; 52 } 53 else 54 { 55 sum += d[ son[u][i] ][0]; 56 flag = 1; 57 } 58 } 59 if(flag) 60 f[u][0] = 0; 61 else 62 f[u][0] = 1; 63 d[u][0] = max(d[u][0], sum); 64 } 65 int main() 66 { 67 int n,m; 68 char worker[110],boss[110]; 69 while(scanf("%d", &n) != EOF && n) 70 { 71 for(int i = 0; i < MAX; i++) 72 son[i].clear(); 73 name.clear(); 74 memset(d, 0, sizeof(d)); 75 memset(f, 0, sizeof(f)); 76 m = 0; 77 scanf("%s", boss); 78 name[boss] = m++; 79 for(int i = 1; i < n; i++) 80 { 81 scanf("%s%s",worker,boss); 82 if(name.count(worker) == 0) //注意输入的处理 83 name[worker] = m++; 84 if(name.count(boss) == 0) 85 name[boss] = m++; 86 son[ name[boss] ].push_back( name[worker] ); 87 } 88 DP(0); 89 if(d[0][0] > d[0][1]) 90 { 91 printf("%d ", d[0][0]); 92 if(f[0][0]) 93 printf("Yes\n"); 94 else 95 printf("No\n"); 96 } 97 else if(d[0][0] < d[0][1]) 98 { 99 printf("%d ", d[0][1]); 100 if(f[0][1]) 101 printf("Yes\n"); 102 else 103 printf("No\n"); 104 } 105 else 106 { 107 printf("%d ", d[0][0]); 108 printf("No\n"); 109 } 110 } 111 return 0; 112 }