UVA 11294 Wedding 2sat

训练指南上的翻译错了,只有一对新娘新郎,其他是夫妻。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\in1.txt","r",stdin);
    freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}


const int maxn=66;
struct TwoSat
{
    int n;
    int mark[maxn*2];
    vector<int> g[maxn*2];
    int s[maxn],c;

    void init(int n)
    {
        this->n=n;
        memset(mark,0,sizeof(mark));
        for(int i=0;i<2*n;i++)
            g[i].clear();
    }
    void add_clause(int x,int xval,int y,int yval)
    {
        x=x*2+xval;
        y=y*2+yval;
        g[x].push_back(y^1);
        g[y].push_back(x^1);
    }
    bool dfs(int u)
    {
        if(mark[u^1])return false ;
        if(mark[u])return true;
        mark[u]=1;
        s[c++]=u;
        for(int i=0;i<g[u].size();i++)
            if(!dfs(g[u][i]))return false;
        return true;
    }
    bool solve()
    {
        for(int i=0;i<2*n;i+=2)
        {
            if(!mark[i]&&!mark[i+1])
            {
                c=0;
                if(!dfs(i))
                {
                    while(c)mark[s[--c]]=0;
                    if(!dfs(i+1))return false;
                }
            }
        }
        return true;
    }
}solver;

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
    {
        solver.init(2*n);
        for(int i=0;i<n;i++)
        {
            solver.add_clause(i*2,0,i*2+1,0);
            solver.add_clause(i*2,1,i*2+1,1);
        }
        int flag=1;
        for(int i=0;i<m;i++)
        {
            char a[5],b[5];
            scanf("%s%s",a,b);
            int j=0,k=0;
            while(a[j]<='9'&&a[j]>='0')j++;
            while(b[k]<='9'&&b[k]>='0')k++;
            int x=0,y=0;
            for(int d=0;d<j;d++)
                x=x*10+a[d]-'0';
            for(int d=0;d<k;d++)
                y=y*10+b[d]-'0';

            x=x*2+(a[j]=='w'?1:0);
            y=y*2+(b[k]=='w'?1:0);
            //solver.add_clause(x,0,y,0);
            solver.add_clause(x,1,y,1);
        }

        solver.mark[1]=1;
        solver.mark[2]=1;
        flag=solver.solve();
        if(!flag)
            printf("bad luck
");
        else
        {
            for(int i=1;i<n;i++)
            {
                printf("%d%c%c",i,solver.mark[i*4]?'h':'w',i==n-1?'
':' ');
            }
        }
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/BMan/p/3619313.html