ZOJ3378 两点间的必经之路 tarjan算法

Attack the NEET Princess

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Time Limit: 9 Seconds      Memory Limit:65536 KB

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Houraisan Kaguya (蓬萊山輝夜) is theNEET (not in education, employment or training) princess living in Eientei(永遠亭). She is an unemployed geek, who sits at home andsurfs the Internet all day. But today is different, as the Hakurei ShrineReitaisai Festival (博麗神社例大祭) will be hosted in HakureiShrine, Kaguya decides to go to buy some new doujin games.

Fujiwara no Mokou (藤原妹紅) detestsKaguya, and has been planing to attack her for a long time. But she nerver gotthe chance before because Kaguya kept staying in Eientei, under theprotection of Yagokoro Eirin (八意永琳). Knowing thatKaguya will go outside today, Mokou decides to wait on certain road and attackher once she passes there. Mokou don't know which path Kaguya will choose, butshe thinks that there always exits some roads where she can always meet Kaguya.

Input

There are multiple cases. Each case begins with two integers 2 ≤ n≤ 10000 -- the number of villages and 2 ≤ m ≤ 100000 -- the number ofroads. Then m lines, each contains two integers a and b (0≤ a, b < n), indicating a road connecting village a andvillage b. Village 0 is Eientei and Village n-1 is HakureiShrine. They are always connected by roads. There may be more than oneroads between two villages.

Output

Find out the roads where Mokou can always meet Kaguya. For each case,output the number of roads in the first line, then output the numbers of theroads in ascending order in the second line.

Sample Input

3 2

0 1

1 2

 

7 8

0 1

0 2

1 3

2 3

3 4

3 5

4 6

5 6

 

Sample Output

2

0 1

0

 

 

求起点到终点之间的必经之路,思路是:首先用tarjan算法求出图中的所有割边,然后从起点开始BFS到终点搜出一条路径,把路径上的边和已求出的割边求交集就可以得出必经之路.

 

代码如下:

 #include<stdio.h>#include<string.h>#include<stack>#define MAXN 10005#include<algorithm>using std::sort;using namespace std;struct E{ int nod,db,cut,num; E *next;}head[MAXN*10];int low[MAXN],dfn[MAXN],res[10*MAXN],cnt;bool mark[MAXN],vis[MAXN];stack <int> ss;void tarjan(int u){ int v; E *p,*s,*q; low[u]=dfn[u]=cnt++; ss.push(u); vis[u]=1; mark[u]=1; for (p=head[u].next;p!=NULL;p=p->next) { v=p->nod; if (!mark[v]) { if (!p->db) { s=head[v].next; q=&head[v]; while (s->nod!=u) {q=s;s=s->next;} q->next=s->next; delete(s); } mark[v]=1; vis[v]=1; ss.push(v); tarjan(v); low[u]=min(low[u],low[v]); } else if (vis[v]) low[u]=min(low[u],dfn[v]); } if (low[u]==dfn[u]) { while (!ss.empty()&&ss.top()!=u) ss.pop(); ss.pop(); }}bool DFS(int b,int e){ E *p; int v; if (b==e) return 1; p=head[b].next; while (p!=NULL) { v=p->nod; if (!mark[v]) { if (DFS(v,e)==1) { if (p->cut==1) p->cut=2; return 1; } } } return 0;}bool find(int a,int b){ E *p; p=head[a].next; while (p!=NULL) { if (p->nod==b) return 1; p=p->next; } return 0;}int main(){ int n,m,i,a,b,total; E *s,*p; while (scanf("%d%d",&n,&m)!=EOF) { memset(mark,0,sizeof(mark)); memset(vis,0,sizeof(vis)); memset(head,0,sizeof(head)); for (i=1;i<=m;++i) { scanf("%d%d",&a,&b); a++;b++; if (!find(a,b)) { s=new E; s->nod=b; s->db=0; s->cut=0; s->num=i; s->next=head[a].next; head[a].next=s; s=new E; s->nod=a; s->db=0; s->cut=0; s->num=i; s->next=head[b].next; head[b].next=s; } else { s=head[a].next; while (s!=NULL&&s->nod!=b) s=s->next; s->db=1; s=head[b].next; while (s!=NULL&&s->nod!=a) s=s->next; s->db=1; } } cnt=1; tarjan(1); for (i=1;i<=n;++i) { p=head[i].next; while (p!=NULL) { if (dfn[i]<low[p->nod]) p->cut=1; p=p->next; } } memset(mark,0,sizeof(mark)); DFS(1,n); total=0; for (i=1;i<=n;++i) { p=head[i].next; while (p!=NULL) { if (p->cut==2) res[++total]=p->num; p=p->next; } } sort(res+1,res+total+1); printf("%d/n",total); for (i=1;i<=total;++i) { printf("%d",res[i]-1); if (i!=total) printf(" "); else printf("/n"); } for (i=1;i<=n;++i) { p=head[i].next; s=p; while (s!=NULL) { s=s->next; delete(p); p=s; } } } return 0;}