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;
}