POJ 3177 Redundant Paths
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12598 | Accepted: 5330 |
Description
Given a description of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way.
There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.
Input
Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.
Output
Sample Input
7 7
1 2
2 3
3 4
2 5
4 5
5 6
5 7
Sample Output
2
Hint
One visualization of the paths is:
1 2 3
+---+---+
| |
| |
6 +---+---+ 4
/ 5
/
/
7 +
Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions.
1 2 3
+---+---+
: | |
: | |
6 +---+---+ 4
/ 5 :
/ :
/ :
7 + - - - -
Check some of the routes:
1 – 2: 1 –> 2 and 1 –> 6 –> 5 –> 2
1 – 4: 1 –> 2 –> 3 –> 4 and 1 –> 6 –> 5 –> 4
3 – 7: 3 –> 4 –> 7 and 3 –> 2 –> 5 –> 7
Every pair of fields is, in fact, connected by two routes.
It's possible that adding some other path will also solve the problem (like one from 6 to 7). Adding two paths, however, is the minimum.
/*这是一个63分的代码，因为没有注意到题目中的重边问题，以后要注意有重边的图和没有重边的图的tarjan求桥的算法，是不同的*/
#include<iostream>
using namespace std;
#include<cstdio>
#define N 5001
#define R 10010
#include<stack>
#include<queue>
#include<cstring>
queue<int>que;
bool qiao[R]={},visit[N]={},visit_edge[R<<];
struct Edge{
int u,v,last;
}edge[R*];
int head[N],du[N],f,r,father[N],dfn[N],low[N],topt=,t=-;
int ans[N]={};
void add_edge(int u,int v)
{
++t;
edge[t].u=u;
edge[t].v=v;
edge[t].last=head[u];
head[u]=t;
}
void input()
{
memset(head,-,sizeof(head));
int u,v;
scanf("%d%d",&f,&r);
for(int i=;i<=r;++i)
{
scanf("%d%d",&u,&v);
add_edge(u,v);
add_edge(v,u);
}
r<<=;
}
void tarjan(int u)
{
dfn[u]=low[u]=++topt;
for(int l=head[u];l!=-;l=edge[l].last)
{
int v=edge[l].v;
if(!visit_edge[l]&&!visit_edge[l^])
{
visit_edge[l]=true;
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
if(low[v]>dfn[u]) qiao[l]=true;
}
else low[u]=min(low[u],dfn[v]);
}
}
}
void suo_dian()
{
for(int i=;i<=f;++i)
{
if(!visit[i])
{
ans[++ans[]]=i;
que.push(i);
visit[i]=true;
while(!que.empty())
{
int x=que.front();
father[x]=i;
que.pop();
for(int l=head[x];l!=-;l=edge[l].last)
{
if(qiao[l]||visit[edge[l].v]) continue;
que.push(edge[l].v);
visit[edge[l].v]=true;
}
} }
}
}
void re_jiantu()
{
for(int l=;l<=r;++l)
{
if(father[edge[l].u]!=father[edge[l].v])
{
du[father[edge[l].u]]++;
du[father[edge[l].v]]++;
}
}
}
int main()
{
freopen("rpaths.in","r",stdin);
freopen("rpaths.out","w",stdout);
input();
for(int i=;i<=f;++i)
{
if(!dfn[i])
tarjan(i);
}
suo_dian();
re_jiantu();
int sum=;
for(int i=;i<=ans[];++i)
if(du[ans[i]]==)
sum++;
printf("%d\n",(sum+)/);
fclose(stdin);fclose(stdout);
return ;
}
正确代码及模板：
#define N 5011
#include<iostream>
using namespace std;
#define M 10010
#include<cstdio>
#include<cstring>
struct Gra{
int n,m,ans,head[N],topt,dfn[N],low[N],t,cnt[N];
bool visit[M<<];
struct Edge{
int v,last;
}edge[M<<];
void init(int f,int r)
{/*初始化不要在上面，上面只是声明，不是变量*/
ans=,topt=,t=-;
n=f;m=r;
memset(head,-,sizeof(head));
memset(dfn,,sizeof(dfn));
memset(low,,sizeof(low));
memset(cnt,,sizeof(cnt));
memset(visit,false,sizeof(visit));
}
void add_edge(int x,int y)
{
++t;
edge[t].v=y;
edge[t].last=head[x];
head[x]=t;
}
void tarjan(int u)
{
dfn[u]=low[u]=++topt;
for(int l=head[u];l!=-;l=edge[l].last)
{
if(visit[l]) continue;
visit[l]=visit[l^]=true;/*找到无向边拆成的另一条边*/
int v=edge[l].v;
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[v],low[u]);
}
else low[u]=min(low[u],dfn[v]);/*多次返祖*/
}
}
void start()
{
for(int i=;i<=n;++i)
if(!dfn[i])
tarjan(i);
for(int i=;i<=n;++i)/*处理缩点以后的图*/
for(int l=head[i];l!=-;l=edge[l].last)
{
int v=edge[l].v;
if(low[i]!=low[v])
cnt[low[v]]++;
/*low[x]!=low[y]说明从low[y]回不到low[x]，那么low[x]--low[y]是一条桥，因为tarjan中多次返祖*/
}
for(int i=;i<=n;++i)
if(cnt[i]==) ans++;/*统计度数是1的叶子节点的数目*/
printf("%d\n",(ans+)>>);
}
}G;
int main()
{
int n,m;
scanf("%d%d",&n,&m);
G.init(n,m);
int x,y;
for(int i=;i<=m;++i)
{
scanf("%d%d",&x,&y);
G.add_edge(x,y);
G.add_edge(y,x);
}
G.start();
return ;
}