标签：ctime include space 1.0 int idt als pre possible
题目链接：
|
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 the link entirely inside the circle or entirely outside the circle. Now liympanda tells Ikki no two links touch inside/outside the circle, except on the boundary. He wants Ikki to figure out whether this is possible… 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 ai and bi, which denote the endpoints of the ith wire. Every point will have at most one link. Output Output a line, either “ Sample Input 4 2 0 1 3 2 Sample Output panda is telling the truth... Source
POJ Monthly--2007.03.04, Ikki
|
[Submit] [Go Back] [Status] [Discuss]
题目意思：n个点。标号为0~n-1,顺序摆在一个圆盘的边缘上，如今给出m条边，每条边连接两个点。能够在圆盘里面或外面。推断这些边是否交叉。
解题思路：
2-SAT
把边抽象成点，假设内部连成的边抽象为i。则相应的边在外部则为i‘,推断随意两条边是否冲突。假设冲突的话。仅仅能一个放在外面。一个放在里面。
i和j冲突，则建图i‘->j j->i‘ i->j‘ j‘->i
代码：
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 1100
int n,m;
vector<vector<int> >myv;
int low[Maxn],dfn[Maxn],sta[Maxn],bc,sc,dep;
int in[Maxn];
bool iss[Maxn];
int x[Maxn],y[Maxn];
bool iscan(int a,int b)
{
if(x[a]<x[b]&&(y[a]>x[b]&&y[a]<y[b]))
return true;
if(x[a]>x[b]&&x[a]<y[b]&&y[a]>y[b])
return true;
return false;
}
void tarjan(int cur)
{
int ne;
low[cur]=dfn[cur]=++dep;
sta[++sc]=cur;
iss[cur]=true;
for(int i=0;i<myv[cur].size();i++)
{
ne=myv[cur][i];
if(!dfn[ne])
{
tarjan(ne);
low[cur]=min(low[cur],low[ne]);
}
else if(iss[ne]&&dfn[ne]<low[cur])
low[cur]=dfn[ne];
}
if(low[cur]==dfn[cur])
{
++bc;
do
{
ne=sta[sc--];
in[ne]=bc;
iss[ne]=false;
}while(ne!=cur);
}
}
void solve()
{
low[1]=dfn[1]=bc=sc=dep=0;
memset(iss,false,sizeof(iss));
memset(dfn,0,sizeof(dfn));
for(int i=1;i<=2*m;i++)
if(!dfn[i])
tarjan(i);
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(~scanf("%d%d",&n,&m))
{
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x[i],&y[i]);
if(x[i]>y[i])
swap(x[i],y[i]);
}
myv.clear();
myv.resize(m*2+10);
for(int i=1;i<=m;i++)
{
for(int j=i+1;j<=m;j++)
{
if(iscan(i,j))
{
myv[2*i].push_back(2*j-1);
myv[2*j-1].push_back(2*i);
myv[2*i-1].push_back(2*j);
myv[2*j].push_back(2*i-1);
}
}
}
solve();
bool ans=true;
for(int i=1;i<=m;i++)
{
if(in[2*i]==in[2*i-1])
{
ans=false;
break;
}
}
if(ans)
printf("panda is telling the truth...\n");
else
printf("the evil panda is lying again\n");
}
return 0;
}
[2-SAT] poj 3207 Ikki's Story IV - Panda's Trick
标签：ctime include space 1.0 int idt als pre possible
原文地址：http://www.cnblogs.com/claireyuancy/p/6888848.html
