pku 3207 Ikki's Story IV - Panda's Trick 2-sat判定是否存在可行解

http://poj.org/problem?id=3207

题意：

一个圆盘的边沿上有n个点, 下标从0开始, 有m条线连接2m个互不相同的点, 线可以在圆盘内部,也可以在圆盘外部, 要求任意两条线不能相交. 给出m条线(内外随意), 问是否满足每条线都不相交.

思路：

可以将第i条线看成一对顶点,编号分别为2*i和2*i+1.那么如果线段i与j相交,就在2*i与2*j+1以及2*i+1与2*j之间连一条双向边。然后就转化到2-sat上判断是否存在可行解了。

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define N 1007
#define M 1007
using namespace std;

struct side
{
    int s,e;
}sid[M];

int n,m;
struct node
{
    int v;
    int next;
}g[M*M];
int head[M],ct;

int dfn[M],low[M];
int belong[M],stk[M];
bool isn[M];
int idx,cnt,top;



bool isok(int i,int j)
{
    //判断相交，建图是关键
    if ((sid[i].s < sid[j].s && sid[i].e > sid[j].s && sid[i].e < sid[j].e)
        || (sid[i].s > sid[j].s && sid[i].s < sid[j].e && sid[i].e > sid[j].e))
        return true;
    else return false;
}
void add(int u,int v)
{
    g[ct].v = v;
    g[ct].next = head[u];
    head[u] = ct++;

    g[ct].v = u;
    g[ct].next = head[v];
    head[v] = ct++;
}
void tarjan(int u)
{
    int i,j;
    dfn[u] = low[u] = ++idx;
    stk[++top] = u;
    isn[u] = true;
    for (i = head[u]; i != - 1; i = g[i].next)
    {
        int v = g[i].v;
        if (dfn[v] == -1)
        {
            tarjan(v);
            low[u] = min(low[u],low[v]);
        }
        else if (isn[v])
        {
            low[u] = min(low[u],dfn[v]);
        }
    }
    if (dfn[u] == low[u])
    {
        cnt++;
        do
        {
            j = stk[top--];
            belong[j] = cnt;
            isn[j] = false;
        }while (j != u);
    }
}

void solve()
{
    int i;
    for (i = 0; i < 2*m; ++i)
    {
        dfn[i] = low[i] = -1;
        isn[i] = false;
        belong[i] = 0;
    }
    idx = cnt = top = 0;

    for (i = 0; i < 2*m; ++i)
    {
        if (dfn[i] == -1) tarjan(i);
    }
    bool flag = false;
    for (i = 0; i < m; ++i)
    {
        if (belong[2*i] == belong[2*i + 1])
        {
            flag = true;
            break;
        }
    }
    if (flag) printf("the evil panda is lying again\n");
    else printf("panda is telling the truth...\n");

}
int main()
{
   // Read();
    int i,j;
    while (~scanf("%d%d",&n,&m))
    {
        for (i = 0; i < m; ++i)
        {
            scanf("%d%d",&sid[i].s,&sid[i].e);
            if (sid[i].s > sid[i].e)
            swap(sid[i].s,sid[i].e);
        }
        CL(head,-1); ct = 0;
        for (i = 0; i < m; ++i)
        {
            for (j = i + 1; j < m; ++j)
            {
                if (isok(i,j))
                {
                    add(2*i,2*j + 1);
                    add(2*j,2*i + 1);
                }
            }
        }
        solve();
    }
    return 0;
}