额。。。我不能很好的解释这个算法，不过有一篇博客讲得挺好的，就是可能有点啰嗦。。。放在下面：

http://blog.csdn.net/moguxiaozhe/article/details/49047085

然后我一开始不会写，问副校长yf要了份代码，发现他写的好优美啊，果断学习了一波，这是我的代码，包括了o(n+m)的tarjan缩点+拓扑算法和o(nm)的暴力算法：

//o(n+m)
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int N=16010,M=40010;
int n,m,mark[N];
struct E{int to,nxt;};
struct Graph
{
    E e[M];
    int head[N],cnt;
    void init(){memset(head,0,sizeof head);cnt=0;}
    void addedge(int x,int y){e[++cnt]=(E){y,head[x]};head[x]=cnt;}
};
struct twosat
{
    Graph G1,G2;
    int n,dfn[N],low[N],cnt,ind,in[N],s[N],top,col[N],bel[N],vis[N];
    bool ins[N];
    vector<int>vec[N];
    void init(int n)
    {
        this->n=n; G1.init(); G2.init();
        memset(dfn,0,sizeof dfn);
        memset(low,0,sizeof low);
        memset(in,0,sizeof in);
        memset(col,0,sizeof col);
        for (int i=0;i<2*n;i++) vec[i].clear();
        cnt=ind=0;
    }
    void add(int x,int y){G1.addedge(x,y^1); G1.addedge(y,x^1);}
    void tarjan(int u)
    {
        dfn[u]=low[u]=++ind; s[++top]=u; ins[u]=1;
        for (int i=G1.head[u];i;i=G1.e[i].nxt)
        {
            int v=G1.e[i].to;
            if (!dfn[v]) tarjan(v),low[u]=min(low[u],low[v]);
            else if (ins[v]) low[u]=min(low[u],dfn[v]);
        }
        if (dfn[u]==low[u])
        {
            while (1)
            {
                int x=s[top--];
                bel[x]=cnt;
                vec[cnt].push_back(x);
                ins[x]=0;
                if (x==u) break;
            }
            cnt++;
        }
    }
    bool work()
    {
        for (int i=0;i<2*n;i++) if (!dfn[i]) tarjan(i);
        for (int i=0;i<2*n;i+=2) if (bel[i]==bel[i^1]) return 0;
        queue<int>Q;
        for (int u=0;u<2*n;u++)
            for (int i=G1.head[u];i;i=G1.e[i].nxt)
            {
                int v=G1.e[i].to;
                if (bel[u]!=bel[v]) G2.addedge(bel[v],bel[u]),in[bel[u]]++;
            }
        for (int i=0;i<cnt;i++) if (!in[i]) Q.push(i);
        while (!Q.empty())
        {
            int u=Q.front(); Q.pop();
            if (!col[u])
            {
                col[u]=1;
                for (int i=0;i<vec[u].size();i++) col[bel[vec[u][i]^1]]=2;
            }
            for (int i=G2.head[u];i;i=G2.e[i].nxt)
            {
                int v=G2.e[i].to;
                if (--in[v]==0) Q.push(v);
            }
        }
        for (int i=0;i<2*n;i++) mark[i]=col[bel[i]];
        return 1;
    }
}A;
int read() {int d=0,f=1; char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();} while (c>='0'&&c<='9') d=(d<<3)+(d<<1)+c-48,c=getchar(); return d*f;}
int main()
{
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        A.init(n);
        for (int i=1;i<=m;i++) A.add(read()-1,read()-1);
        if (!A.work()) {puts("NIE"); continue;}
        for (int i=0;i<2*n;i++)
            if (mark[i]==1) printf("%d\n",i+1);
    }
    return 0;
}
//o(n*m)
#include<cstdio>
#include<cstring>
const int N=160100,M=400000;
struct edge{int to,nxt;}e[M<<1];
int n,m,head[N],cnt,s[N],top;
bool b[N];
int read() {int d=0,f=1; char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();} while (c>='0'&&c<='9') d=(d<<3)+(d<<1)+c-48,c=getchar(); return d*f;}
void addedge(int x,int y){e[++cnt]=(edge){y,head[x]}; head[x]=cnt;}
bool dfs(int u)
{
    if (b[u^1]) return 0;
    if (b[u]) return 1;
    b[u]=1; s[++top]=u;
    for (int i=head[u];i;i=e[i].nxt)
    {
        int v=e[i].to;
        if (!dfs(v)) return 0;
    }
    return 1;
}
int main()
{
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        memset(head,0,sizeof head);
        memset(b,0,sizeof b);
        cnt=0;
        for (int i=1;i<=m;i++)
        {
            int x=read()-1,y=read()-1;
            addedge(x,y^1); addedge(y,x^1);
        }
        bool bo=1;
        for (int i=0;i<2*n;i+=2)
            if (!b[i]&&!b[i+1])
            {
                top=0;
                if (!dfs(i))
                {
                    while (top) b[s[top--]]=0;
                    if (!dfs(i+1)) {bo=0; break;}
                }
            }
        if (!bo) puts("NIE");
        else
            for (int i=0;i<2*n;i++)
                if (b[i]) printf("%d\n",i+1);
    }
    return 0;
}
View Code

据周大神说这个o(nm)是假的，有大佬曾试图证明它是线形的，结果有一个可以卡成o(n^2)的数据，不过n^2还是很优秀的~

接下来做了些题，poj3207发现一直RE，原来是我边数组开小了。。。

#include<cstdio>
#include<iostream>
using namespace std;
const int N=2017;
struct E{int to,nxt;}e[N*N];
int n,m,a[N],b[N],dfn[N],low[N],ind,cnt,head[N],scc[N],s[N],top;
bool ins[N];
void addedge(int x,int y)
{
    e[++cnt]=(E){y,head[x]}; head[x]=cnt;
    e[++cnt]=(E){x,head[y]}; head[y]=cnt;
}
void tarjan(int u)
{
    dfn[u]=low[u]=++ind; s[++top]=u; ins[u]=1;
    for (int i=head[u];i;i=e[i].nxt)
    {
        int v=e[i].to;
        if (!dfn[v]) tarjan(v),low[u]=min(low[u],low[v]);
        else if (ins[v]) low[u]=min(low[u],dfn[v]);
    }
    if (dfn[u]==low[u])
    {
        cnt++;
        while (1)
        {
            int x=s[top--];
            ins[x]=0; scc[x]=cnt;
            if (x==u) break;
        }
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d",&a[i],&b[i]);
        if (a[i]>b[i]) swap(a[i],b[i]);
    }
    for (int i=1;i<=m;i++)
        for (int j=1;j<i;j++)
            if ((a[i]<a[j]&&b[i]>a[j]&&b[i]<b[j])||(a[i]>a[j]&&b[j]>a[i]&&b[i]>b[j]))
            {
                addedge(i<<1,j<<1|1);
                addedge(i<<1|1,j<<1);
            }
    cnt=0;
    for (int i=1;i<=m*2;i++) if (!dfn[i]) tarjan(i);
    for (int i=1;i<=m;i++)
        if (scc[i<<1]==scc[i<<1|1])
        {
            puts("the evil panda is lying again");
            return 0;
        }
    puts("panda is telling the truth...");
    return 0;
}
View Code

然后是bzoj1823，大概题意是至少选定a,b中的一个，这样我们只要连a'->b,b'->a，表示选了a'就必须选b，就好了。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<string>
#include<queue>
#include<map>
#include<set>
#include<vector>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define sqr(x) (x)*(x)
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define per(i,x,y) for (int i=(x);i>=(y);i--)
using namespace std;
typedef long long LL;
typedef double DBD;
typedef pair<int,int> pa;
const int inf=1e9;
const LL INF=1e18;
//-----------------------------------------------head-------------------------------------------//
const int N=211,M=2017;
struct E{int to,nxt;};
struct Graph
{
    E e[M<<1];
    int cnt,head[N];
    void init(){memset(head,0,sizeof head); cnt=0;}
    void addedge(int x,int y){e[++cnt]=(E){y,head[x]};head[x]=cnt;}
};
struct twosat
{
    Graph G1,G2;
    int n,dfn[N],low[N],ind,cnt,scc[N],s[N],top;
    bool ins[N];
    void init(int n)
    {
        this->n=n; G1.init();
        memset(dfn,0,sizeof dfn);
        memset(low,0,sizeof low);
        ind=cnt=0;
    }
    void add(int x,int y){G1.addedge(x,y);}
    void tarjan(int u)
    {
        dfn[u]=low[u]=++ind; s[++top]=u; ins[u]=1;
        for (int i=G1.head[u];i;i=G1.e[i].nxt)
        {
            int v=G1.e[i].to;
            if (!dfn[v]) tarjan(v),low[u]=min(low[u],low[v]);
            else if (ins[v]) low[u]=min(low[u],dfn[v]);
        }
        if (dfn[u]==low[u])
        {
            cnt++;
            while (1)
            {
                int x=s[top--];
                ins[x]=0; scc[x]=cnt;
                if (x==u) break;
            }
        }
    }
    bool work()
    {
        for (int i=2;i<=2*n+1;i++) if (!dfn[i]) tarjan(i);
        for (int i=2;i<=2*n+1;i+=2) if (scc[i]==scc[i+1]) return 0;
        return 1;
    }
}A;
int Write[20];
int n,m,T;
char s1[3],s2[3];
int read() {int d=0,f=1; char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();} while (c>='0'&&c<='9') d=(d<<3)+(d<<1)+c-48,c=getchar(); return d*f;}
void write(int x){int t=0; if (x<0) putchar('-'),x=-x; for (;x;x/=10) Write[++t]=x%10; if (!t) putchar('0'); for (int i=t;i>=1;i--) putchar((char)(Write[i]+48));}
void judge(){freopen(".in","r",stdin); freopen(".out","w",stdout);}
int main()
{
    //judge();
    T=read();
    while (T--)
    {
        n=read(); m=read();
        A.init(n);
        while (m--)
        {
            int b1,b2,x,y;
            char c=getchar();
            while (c!='m'&&c!='h') c=getchar();
            if (c=='m') b1=1; else b1=0;
            x=read();
            c=getchar();
            while (c!='m'&&c!='h') c=getchar();
            if (c=='m') b2=1; else b2=0;
            y=read();
            x<<=1; y<<=1;
            x+=b1; y+=b2;
            int xx=x,yy=y;
            if (b1) xx--; else xx++;
            if (b2) yy--; else yy++;
            A.add(xx,y); A.add(yy,x);
            //printf("%d %d %d %d\n",xx,y,yy,x);
        }
        if (A.work()) puts("GOOD");    else puts("BAD");
    }
    return 0;
}
View Code

还在se题中，感觉2-sat的题主要还是转化模型。