Redundant Paths
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13712 | Accepted: 5821 |
Description
In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another.
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.
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
Line 1: Two space-separated integers: F and R
Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.
Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.
Output
Line 1: A single integer that is the number of new paths that must be built.
Sample Input
7 7 1 2 2 3 3 4 2 5 4 5 5 6 5 7
Sample Output
2
Hint
Explanation of the sample:
One visualization of the paths is:
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.
One visualization of the paths is:
1 2 3Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions.
+---+---+
| |
| |
6 +---+---+ 4
/ 5
/
/
7 +
1 2 3Check some of the routes:
+---+---+
: | |
: | |
6 +---+---+ 4
/ 5 :
/ :
/ :
7 + - - - -
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.
Source
原题大意：问要加几条边能使一个图变为双连通分量。
解题思路：用tarjian将图缩成一棵树，很明显只要将叶子结点二分相连即可，也就是说如果叶子结点数为偶数，答案为叶子节点个数；如果为奇数则+1再整除2.
然后发现无论奇数还是偶数，全部+1整除2就可以了。
#include<stdio.h>
#include<string.h>
struct mp
{
int begin,to,next;
}map[10010];
int frist[10010],num,dfn[10010],low[10010],times,bridgenum,father[10010];
int degree[10010],con[10010],stack[10010],cnt,top,s[10010];
bool count[10010];
void init()
{
memset(father,0,sizeof(father));
memset(count,0,sizeof(count));
memset(con,0,sizeof(con));
memset(degree,0,sizeof(degree));
memset(s,0,sizeof(s));
memset(stack,0,sizeof(stack));
memset(frist,0,sizeof(frist));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(map,0,sizeof(map));
num=times=bridgenum=cnt=top=0;
}
void add(int x,int y)
{
++num;
map[num].begin=x;map[num].to=y;
map[num].next=frist[x];frist[x]=num;
return;
}
void tarjian(int v)
{
int i;bool flag=true;
dfn[v]=low[v]=++times;
stack[top++]=v;
for(i=frist[v];i;i=map[i].next)
{
if(map[i].to==father[v]&&flag)
{
flag=false;
continue;
}
if(!dfn[map[i].to])
{
father[map[i].to]=v;
tarjian(map[i].to);
if(low[map[i].to]<low[v]) low[v]=low[map[i].to];
}
else
if(dfn[map[i].to]<low[v]) low[v]=dfn[map[i].to];
}
if(low[v]==dfn[v])
{
++cnt;
do
{
v=stack[--top];
s[v]=cnt;
}while(dfn[v]!=low[v]);
}
}
int main()
{
int n,m,a,b,i,u,v,k=0,ans=0,j;
init();
scanf("%d%d",&n,&m);
for(i=0;i<m;++i)
{
scanf("%d%d",&a,&b);
add(a,b);add(b,a);
}
tarjian(1);
for(i=1;i<=n;++i)
{
for(j=frist[i];j;j=map[j].next)
{
v=map[j].to;
if(s[i]!=s[v])
{
++degree[s[i]];
++degree[s[v]];
}
}
}
for(i=1;i<=cnt;++i)
if(degree[i]==2)
++ans;
printf("%d
",(ans+1)/2);
return 0;
}