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Organising a group trip for the elderly can be a daunting task... Not least because of the fussy participants, each of whom will only make the trip on condition that some other participant also comes.
After some effort,you have taken from each of your participants a number, indicating that this participant will refuse to join the excursion unless the participant with that number also joins– the less choosy simply give their own number. This would be easy enough to resolve (just send all of them) but the bus you are going to use during the trip has only a ?xed number of places.
Task
Given the preferences of all participants, ?nd the maximum number of participants that can join.
Input
The ?rst line of input contains two integers n and k (1 ≤ k ≤ n ≤ 1000), where n denotes the total number of participants and k denotes the number of places on the bus. The second line contains n integers xi for i = 1,2,...,n, where 1 ≤ xi ≤ n. The meaningof xi is that the i-th participant will refuse to join the excursion unless the xi-th participant also joins.
Output
Output one integer: the maximum number of participants that can join the excursion, so that all the participants’ preferences are obeyed and the capacity of the bus is not exceeded.
Sample Input 1
4 4
1 2 3 4
Sample Output 1
4
Sample Input 2
12 3
2 3 4 5 6 7 4 7 8 8 12 12
Sample Output 2
2
Sample Input 3
5 4
2 3 1 5 4
Sample Output 3
3
解题：强连通缩点+虚拟根节点+树形dp
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1010; 4 struct arc { 5 int to,next; 6 arc(int x = 0,int y = -1) { 7 to = x; 8 next = y; 9 } 10 } e[maxn<<1]; 11 int head[maxn],low[maxn],dfn[maxn],belong[maxn],rs[maxn],tot,idx,scc; 12 bool instack[maxn],connect[maxn][maxn],dp[maxn][maxn]; 13 int root,n,m,ind[maxn]; 14 vector<int>g[maxn]; 15 stack<int>stk; 16 void tarjan(int u) { 17 dfn[u] = low[u] = ++idx; 18 instack[u] = true; 19 stk.push(u); 20 for(int i = head[u]; ~i; i = e[i].next) { 21 if(!dfn[e[i].to]) { 22 tarjan(e[i].to); 23 low[u] = min(low[u],low[e[i].to]); 24 } else if(instack[e[i].to]) low[u] = min(low[u],dfn[e[i].to]); 25 } 26 if(low[u] == dfn[u]) { 27 int v; 28 scc++; 29 do { 30 instack[v = stk.top()] = false; 31 belong[v] = scc; 32 rs[scc]++; 33 stk.pop(); 34 } while(v != u); 35 } 36 } 37 void add(int u,int v) { 38 e[tot] = arc(v,head[u]); 39 head[u] = tot++; 40 } 41 void init() { 42 while(!stk.empty()) stk.pop(); 43 for(int i = 0; i < maxn; ++i) { 44 dfn[i] = low[i] = 0; 45 rs[i] = belong[i] = 0; 46 instack[i] = false; 47 head[i] = -1; 48 memset(dp[i],false,sizeof dp[i]); 49 memset(connect[i],false,sizeof connect[i]); 50 g[i].clear(); 51 ind[i] = 0; 52 } 53 tot = idx = scc = 0; 54 rs[root = 0] = 0; 55 } 56 void dfs(int u) { 57 dp[u][0] = dp[u][rs[u]] = true; 58 for(int i = g[u].size()-1; i >= 0; --i) { 59 int v = g[u][i]; 60 dfs(v); 61 for(int j = m; j >= rs[u]; --j) { 62 if(dp[u][j]) continue; 63 for(int k = 0; k <= j - rs[u]; ++k) 64 if(dp[v][k] && dp[u][j - k]) { 65 dp[u][j] = true; 66 break; 67 } 68 } 69 } 70 } 71 int main() { 72 int u; 73 while(~scanf("%d %d",&n,&m)) { 74 init(); 75 for(int i = 1; i <= n; ++i) { 76 scanf("%d",&u); 77 add(u,i); 78 } 79 for(int i = 1; i <= n; ++i) 80 if(!dfn[i]) tarjan(i); 81 for(int i = 1; i <= n; ++i) { 82 for(int j = head[i]; ~j; j = e[j].next) { 83 if(belong[i] == belong[e[j].to]) continue; 84 if(connect[belong[i]][belong[e[j].to]]) continue; 85 g[belong[i]].push_back(belong[e[j].to]); 86 ind[belong[e[j].to]]++; 87 connect[belong[i]][belong[e[j].to]] = true; 88 } 89 } 90 for(int i = 1; i <= scc; ++i) 91 if(!ind[i]) g[root].push_back(i); 92 dfs(root); 93 int ret = 0; 94 for(int i = m; i >= 0; --i) 95 if(dp[root][i]) { 96 ret = i; 97 break; 98 } 99 printf("%d\n",ret); 100 } 101 return 0; 102 }
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原文地址：http://www.cnblogs.com/crackpotisback/p/4461729.html
