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;

}

  

 

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