poj3177——Redundant Paths（双连通分量）

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.
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.
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

加最少的边从连通图变成双连通图，有一个结论是结果为（强连通分量出度为1的点个数+1）/2。
题目中还有重边，但前向星是不能判重边，所以要在算法中判断，但判断方法我看这么久，还是不能释怀。。。

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>//#include <map>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 10005#define Mod 10001using namespace std;struct Edge{    int to,next;} edge[MAXN*2];int head[MAXN],tot;int low[MAXN],dfn[MAXN],stack[MAXN],belong[MAXN];int index,top,scc;bool instack[MAXN];int out[MAXN];void addedge(int u,int v){    edge[tot].to=v;    edge[tot].next=head[u];    head[u]=tot++;}void tarjan(int u,int fa){    int v;    low[u]=dfn[u]=++index;    stack[top++]=u;    instack[u]=true;    for(int i=head[u]; i!=-1; i=edge[i].next)    {        v=edge[i].to;        if(i==(fa^1))            continue;        if(!dfn[v])        {            tarjan(v,i);            low[u]=min(low[u],low[v]);        }        else if(instack[v])            low[u]=min(low[u],dfn[v]);    }    if(low[u]==dfn[u])    {        scc++;        do        {            v=stack[--top];            instack[v]=false;            belong[v]=scc;        }        while(v!=u);    }}void init(){    tot=index=scc=top=0;    memset(head,-1,sizeof(head));    memset(dfn,0,sizeof(dfn));    memset(instack,false,sizeof(instack));}int main(){    int m,n;    while(~scanf("%d%d",&n,&m))    {        init();        while(m--)        {            int u,v;            scanf("%d%d",&u,&v);            addedge(u,v);            addedge(v,u);        }        tarjan(1,-1); //因为图本来就连通，所以随便定起点就行        memset(out,0,sizeof(out));        for(int i=1; i<=n; ++i)        {            for(int j=head[i]; j!=-1; j=edge[j].next)            {                int v=edge[j].to;                if(belong[i]!=belong[v])                    out[belong[i]]++;            }        }        int ans=0;        for(int i=1; i<=n; ++i)            if(out[i]==1)                ans++;        ans=(ans+1)/2;        printf("%d\n",ans);    }    return 0;}

嗨呀，又看到篇题解，直接用id来标示每条边，这样判重边的方法就比较清晰了

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>//#include <map>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 10005#define Mod 10001using namespace std;struct Edge{    int to,next;    int id;} edge[MAXN*2];int head[MAXN],tot;int low[MAXN],dfn[MAXN],stack[MAXN],belong[MAXN];int index,top,scc;bool instack[MAXN];int out[MAXN];void addedge(int u,int v,int id){    edge[tot].to=v;    edge[tot].id=id;    edge[tot].next=head[u];    head[u]=tot++;}void tarjan(int u,int fa){    int v;    low[u]=dfn[u]=++index;    stack[top++]=u;    instack[u]=true;    for(int i=head[u]; i!=-1; i=edge[i].next)    {        v=edge[i].to;        if(fa==edge[i].id)            continue;        if(!dfn[v])        {            tarjan(v,edge[i].id);            low[u]=min(low[u],low[v]);        }        else if(instack[v])            low[u]=min(low[u],dfn[v]);    }    if(low[u]==dfn[u])    {        scc++;        do        {            v=stack[--top];            instack[v]=false;            belong[v]=scc;        }        while(v!=u);    }}void init(){    tot=index=scc=top=0;    memset(head,-1,sizeof(head));    memset(dfn,0,sizeof(dfn));    memset(instack,false,sizeof(instack));}int main(){    int m,n;    while(~scanf("%d%d",&n,&m))    {        init();        for(int i=1;i<=m;++i)        {            int u,v;            scanf("%d%d",&u,&v);            addedge(u,v,i);            addedge(v,u,i);        }        tarjan(1,-1); //因为图本来就连通，所以随便定起点就行        memset(out,0,sizeof(out));        for(int i=1; i<=n; ++i)        {            for(int j=head[i]; j!=-1; j=edge[j].next)            {                int v=edge[j].to;                if(belong[i]!=belong[v])                    out[belong[i]]++;            }        }        int ans=0;        for(int i=1; i<=n; ++i)            if(out[i]==1)                ans++;        ans=(ans+1)/2;        printf("%d\n",ans);    }    return 0;}