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  • POJ 3207 Ikki's Story IV

    Ikki's Story IV - Panda's Trick
    Time Limit: 1000MS   Memory Limit: 131072K
    Total Submissions: 6914   Accepted: 2565

    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 “panda is telling the truth...” or “the evil panda is lying again”.

    Sample Input

    4 2
    0 1
    3 2

    Sample Output

    panda is telling the truth...

    --------

    把线看做点,建图,twosat

    -----------

    /** head-file **/
    
    #include <iostream>
    #include <fstream>
    #include <sstream>
    #include <iomanip>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <string>
    #include <vector>
    #include <queue>
    #include <stack>
    #include <list>
    #include <set>
    #include <map>
    #include <algorithm>
    
    /** define-for **/
    
    #define REP(i, n) for (int i=0;i<int(n);++i)
    #define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
    #define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
    #define REP_1(i, n) for (int i=1;i<=int(n);++i)
    #define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
    #define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
    #define REP_N(i, n) for (i=0;i<int(n);++i)
    #define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
    #define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
    #define REP_1_N(i, n) for (i=1;i<=int(n);++i)
    #define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
    #define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
    
    /** define-useful **/
    
    #define clr(x,a) memset(x,a,sizeof(x))
    #define sz(x) int(x.size())
    #define see(x) cerr<<#x<<" "<<x<<endl
    #define se(x) cerr<<" "<<x
    #define pb push_back
    #define mp make_pair
    
    /** test **/
    
    #define Display(A, n, m) {                      
        REP(i, n){                                  
            REP(j, m) cout << A[i][j] << " ";       
            cout << endl;                           
        }                                           
    }
    
    #define Display_1(A, n, m) {                    
        REP_1(i, n){                                
            REP_1(j, m) cout << A[i][j] << " ";     
            cout << endl;                           
        }                                           
    }
    
    using namespace std;
    
    /** typedef **/
    
    typedef long long LL;
    
    /** Add - On **/
    
    const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
    const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
    const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
    
    const int MOD = 1000000007;
    const int INF = 0x3f3f3f3f;
    const long long INFF = 1LL << 60;
    const double EPS = 1e-9;
    const double OO = 1e15;
    const double PI = acos(-1.0); //M_PI;
    
    const int maxn=2111;
    const int maxm=2111111;
    int n,m;
    struct EDGENODE{
        int to;
        int next;
    };
    struct TWO_SAT{
        int head[maxn*2];
        EDGENODE edges[maxm*2];
        int edge;
        int n;
        bool mark[maxn*2];
        int S[maxn*2],c;
        void init(int n){
            this->n=n;
            clr(mark,0);
            clr(head,-1);
            edge=0;
        }
        void addedge(int u,int v){
            edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
        }
        // x = xval or y = yval
        void add_clause(int x,int xval,int y,int yval){
            x=x*2+xval;
            y=y*2+yval;
            addedge(x^1,y);
            addedge(y^1,x);
        }
        bool dfs(int x){
            if (mark[x^1]) return false;
            if (mark[x]) return true;
            mark[x]=true;
            S[c++]=x;
            for (int i=head[x];i!=-1;i=edges[i].next)
                if (!dfs(edges[i].to)) return false;
            return true;
        }
        bool solve(){
            for (int i=0;i<n*2;i+=2)
                if (!mark[i]&&!mark[i+1]){
                    c=0;
                    if (!dfs(i)){
                        while (c>0) mark[S[--c]]=false;
                        if (!dfs(i+1)) return false;
                    }
                }
            return true;
        }
    }TwoSAT;
    
    int A[maxn];
    int B[maxn];
    
    int main()
    {
        while (~scanf("%d%d",&n,&m))
        {
            REP(i,m){
                scanf("%d%d",&A[i],&B[i]);
                if (A[i]>B[i]) swap(A[i],B[i]);
            }
            TwoSAT.init(m);
            REP(i,m)
            {
                int fix=A[i];
                int nw=(B[i]-fix+n)%n;
                REP(j,m)
                {
                    if (i==j) continue;
                    int nx=(A[j]-fix+n)%n;
                    int ny=(B[j]-fix+n)%n;
                    if (nx>ny) swap(nx,ny);
                    if ( nx<nw&&ny>nw ){
                        TwoSAT.add_clause(i,0,j,0);
                        TwoSAT.add_clause(i,1,j,1);
                    }
                }
            }
            if (TwoSAT.solve()) puts("panda is telling the truth...");
            else puts("the evil panda is lying again");
        }
        return 0;
    }
    





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  • 原文地址:https://www.cnblogs.com/cyendra/p/3681603.html
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