The Bottom of a Graph
Time Limit: 3000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 2553
64-bit integer IO format: %lld Java class name: Main
We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges. Then G=(V,E) is called a directed graph.
Let n be a positive integer, and let p=(e1,...,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,...,vn+1). Then p is called a path from vertex v1 to vertex vn+1 in G and we say that vn+1 is reachable from v1, writing (v1→vn+1).
Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w. The bottom of a graph is the subset of all nodes that are sinks, i.e., bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}. You have to calculate the bottom of certain graphs.
Input
The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer numbers in the set V={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integer e and, thereafter, epairs of vertex identifiers v1,w1,...,ve,we with the meaning that (vi,wi)∈E. There are no edges other than specified by these pairs. The last test case is followed by a zero.
Output
For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line.
Sample Input
3 3 1 3 2 3 3 1 2 1 1 2 0
Sample Output
1 3 2
Source
解题：求这样的点，它能到的点，那点也可以到它，注意先后顺序。然后求强联通缩点，出度为0的集合，里面的点即为我们求得那些点
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 6000;
18 struct arc {
19 int to,next;
20 arc(int x = 0,int y = -1) {
21 to = x;
22 next = y;
23 }
24 };
25 arc e[maxn*100];
26 int head[maxn],dfn[maxn],low[maxn],belong[maxn],my[maxn];
27 int tot,n,m,top,scc,idx,out[maxn],ans[maxn];
28 bool instack[maxn];
29 void init() {
30 for(int i = 0; i < maxn; ++i) {
31 dfn[i] = low[i] = belong[i] = 0;
32 instack[i] = false;
33 head[i] = -1;
34 out[i] = 0;
35 }
36 scc = tot = idx = top = 0;
37 }
38 void add(int u,int v){
39 e[tot] = arc(v,head[u]);
40 head[u] = tot++;
41 }
42 void tarjan(int u) {
43 dfn[u] = low[u] = ++idx;
44 my[top++] = u;
45 instack[u] = true;
46 for(int i = head[u]; ~i; i = e[i].next) {
47 if(!dfn[e[i].to]) {
48 tarjan(e[i].to);
49 low[u] = min(low[u],low[e[i].to]);
50 } else if(instack[e[i].to]) low[u] = min(low[u],dfn[e[i].to]);
51 }
52 if(dfn[u] == low[u]) {
53 scc++;
54 int v;
55 do {
56 v = my[--top];
57 instack[v] = false;
58 belong[v] = scc;
59 } while(v != u);
60 }
61 }
62 int main() {
63 int u,v;
64 while(~scanf("%d",&n)&&n) {
65 scanf("%d",&m);
66 init();
67 for(int i = 0; i < m; ++i) {
68 scanf("%d%d",&u,&v);
69 add(u,v);
70 }
71 for(int i = 1; i <= n; ++i)
72 if(!dfn[i]) tarjan(i);
73 for(int i = 1; i <= n; ++i)
74 for(int j = head[i]; ~j; j = e[j].next)
75 if(belong[i] != belong[e[j].to]) out[belong[i]]++;
76 int cnt = 0;
77 for(int i = 1; i <= n; ++i)
78 if(!out[belong[i]]) ans[cnt++] = i;
79 if(cnt){
80 for(int i = 0; i < cnt; ++i)
81 printf("%d%c",ans[i],i + 1 == cnt?‘\n‘:‘ ‘);
82 }else puts("");
83 }
84 return 0;
85 }