poj~1904
Description
So the king asked his wizard to find for each of his sons the girl he liked, so that he could marry her. And the king’s wizard did it — for each son the girl that he could marry was chosen, so that he liked this girl and, of course, each beautiful girl had to marry only one of the king’s sons.
However, the king looked at the list and said: “I like the list you have made, but I am not completely satisfied. For each son I would like to know all the girls that he can marry. Of course, after he marries any of those girls, for each other son you must still be able to choose the girl he likes to marry.”
The problem the king wanted the wizard to solve had become too hard for him. You must save wizard’s head by solving this problem.
Input
The last line of the case contains the original list the wizard had made — N different integer numbers: for each son the number of the girl he would marry in compliance with this list. It is guaranteed that the list is correct, that is, each son likes the girl he must marry according to this list.
Output
Sample Input
4 2 1 2 2 1 2 2 2 3 2 3 4 1 2 3 4
Sample Output
2 1 2 2 1 2 1 3 1 4
Hint
这题加和不加输入输出挂差别是超级大的
下面是放的是加输入输出挂的代码
1 #include <cstdio> 2 #include <cstring> 3 #include <string> 4 #include <algorithm> 5 #include <queue> 6 using namespace std; 7 8 const int maxn = 1e6 + 10; 9 int n, m, u, v, tot, top, cnt, flag; 10 int Scan() 11 { 12 int res=0,ch,flag=0; 13 if((ch=getchar())=='-') 14 flag=1; 15 else if(ch>='0'&&ch<='9') 16 res=ch-'0'; 17 while((ch=getchar())>='0'&&ch<='9') 18 res=res*10+ch-'0'; 19 return flag?-res:res; 20 } 21 void Out(int a) 22 { 23 if(a>9) 24 Out(a/10); 25 putchar(a%10+'0'); 26 } 27 struct node { 28 int v, next; 29 } edge[maxn]; 30 int head[maxn], instack[maxn], s[maxn]; 31 int dfn[maxn], low[maxn], belong[maxn]; 32 void init() { 33 tot = cnt = top = flag = 0; 34 memset(s, 0, sizeof(s)); 35 memset(head, -1, sizeof(head)); 36 memset(dfn, 0, sizeof(dfn)); 37 memset(instack, 0, sizeof(instack)); 38 } 39 void add(int u, int v) { 40 edge[tot].v = v; 41 edge[tot].next = head[u]; 42 head[u] = tot++; 43 } 44 void tarjin(int v) { 45 dfn[v] = low[v] = ++flag; 46 instack[v] = 1; 47 s[top++] = v; 48 for (int i = head[v] ; i != -1 ; i = edge[i].next) { 49 int j = edge[i].v; 50 if (!dfn[j]) { 51 tarjin(j); 52 low[v] = min(low[v], low[j]); 53 } else if (instack[j]) low[v] = min(low[v], dfn[j]); 54 } 55 if (dfn[v] == low[v]) { 56 cnt++; 57 int t; 58 do { 59 t = s[--top]; 60 instack[t] = 0; 61 belong[t] = cnt; 62 } while(t != v) ; 63 } 64 } 65 void solve() { 66 for (int i = 1 ; i <= n ; i++) 67 if (!dfn[i]) tarjin(i); 68 } 69 int ans[maxn]; 70 int main() { 71 while(scanf("%d", &n) != EOF) { 72 init(); 73 int num; 74 memset(ans,0,sizeof(ans)); 75 for (int i = 1 ; i <= n ; i++) { 76 num=Scan(); 77 while(num--) { 78 int v; 79 v=Scan(); 80 add(i, n + v); 81 } 82 } 83 for (int i = 1 ; i <= n ; i++) { 84 int v; 85 v=Scan(); 86 add(n + v, i); 87 } 88 for (int i = 1 ; i <= 2*n ; i++) 89 if (!dfn[i]) tarjin(i); 90 for (int i = 1 ; i <= n ; i++) { 91 int k = 0; 92 for (int j = head[i] ; ~j ; j = edge[j].next) { 93 if (belong[edge[j].v] == belong[i]) ans[k++] = edge[j].v - n; 94 } 95 sort(ans, ans + k); 96 Out(k); 97 for (int i = 0 ; i < k ; i++){ 98 putchar(' '); 99 Out(ans[i]); 100 } 101 putchar('\n'); 102 } 103 } 104 return 0; 105 }