Language: Default Ikki's Story IV - Panda's Trick
Description liympanda, one of Ikki’s friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many times, he is tired of such easy games and wants to play another game with Ikki. liympanda has a magic circle and he puts it on a plane, there are n points on its boundary in circular border: 0, 1, 2, …, n − 1. Evil panda claims that he is connecting m pairs of points. To connect two points, liympanda either places Despaired at the minesweeping game just played, Ikki is totally at a loss, so he decides to write a program to help him. Input The input contains exactly one test case. In the test case there will be a line consisting of of two integers: n and m (n ≤ 1,000, m ≤ 500). The following m lines each contain two integers a and b, which denote Output Output a line, either “ Sample Input 4 2 Sample Output panda is telling the truth... Source POJ Monthly--2007.03.04, Ikki |
题意：有n个点在一个圆上。标号0~n-1。有m条线段告诉两个端点的标号。每条线段能够在画在圆内也能够画在圆外。问这m条线段是否能不向交。
思路：2-SAT模板题，把每条线段当做节点。每条线有两种状态（圆内和圆外。两者仅仅能取其一），建图后求强连通，看 i 和 i‘ 是否在同一个连通分量，若在则无解，否则有解。
參考：由对称性解2-SAT问题 还有 kuangbin神牛的总结点击打开链接
代码：
#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std; #define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 2005;
const int MAXM = 200010;
const int N = 1005; struct Edge
{
int to,next;
}edge[MAXM]; int head[MAXN],tot; void init()
{
tot=0;
memset(head,-1,sizeof(head));
} void addedge(int u,int v)
{
edge[tot].to=v;
edge[tot].next=head[u];
head[u]=tot++;
} int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
int Index,scc,top;
bool Instack[MAXN]; void Tarjan(int u)
{
int v;
Low[u]=DFN[u]=++Index;
Stack[top++]=u;
Instack[u]=true;
for (int i=head[u];~i;i=edge[i].next)
{
v=edge[i].to;
if (!DFN[v])
{
Tarjan(v);
if (Low[u]>Low[v])
Low[u]=Low[v];
}
else if (Instack[v]&&Low[u]>DFN[v])
Low[u]=DFN[v];
}
if (Low[u]==DFN[u])
{
++scc;
do{
v=Stack[--top];
Instack[v]=false;
Belong[v]=scc;
}while (v!=u);
}
return ;
} bool solvable(int n)
{
memset(DFN,0,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
Index=scc=top=0;
for (int i=0;i<n;i+=2)
if (!DFN[i])
Tarjan(i);
for (int i=0;i<n;i+=2)
{
if (Belong[i]==Belong[i^1])
return false;
}
return true;
} struct Line
{
int s,t;
}line[MAXN]; int n,m; int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
int i,j;
scanf("%d%d",&n,&m);
init();
for (i=0;i<m;i++)
{
scanf("%d%d",&line[i].s,&line[i].t);
if (line[i].s>line[i].t) swap(line[i].s,line[i].t);
}
for (i=0;i<m;i++)
{
for (j=i+1;j<m;j++)
{
if (line[i].s<line[j].s&&line[i].t>line[j].s&&line[i].t<line[j].t||line[j].s<line[i].s&&line[j].t>line[i].s&&line[j].t<line[i].t)
{
addedge(2*i,2*j+1);
addedge(2*j+1,2*i);
addedge(2*j,2*i+1);
addedge(2*i+1,2*j);
}
}
}
if (solvable(m*2))
printf("panda is telling the truth...\n");
else
printf("the evil panda is lying again\n");
return 0;
}