• HDU-1814 Peaceful Commission 2sat


      题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1814

      简单的2sat题。

      1 //STATUS:C++_AC_390MS_996KB
      2 #include <functional>
      3 #include <algorithm>
      4 #include <iostream>
      5 //#include <ext/rope>
      6 #include <fstream>
      7 #include <sstream>
      8 #include <iomanip>
      9 #include <numeric>
     10 #include <cstring>
     11 #include <cassert>
     12 #include <cstdio>
     13 #include <string>
     14 #include <vector>
     15 #include <bitset>
     16 #include <queue>
     17 #include <stack>
     18 #include <cmath>
     19 #include <ctime>
     20 #include <list>
     21 #include <set>
     22 #include <map>
     23 using namespace std;
     24 //using namespace __gnu_cxx;
     25 //define
     26 #define pii pair<int,int>
     27 #define mem(a,b) memset(a,b,sizeof(a))
     28 #define lson l,mid,rt<<1
     29 #define rson mid+1,r,rt<<1|1
     30 #define PI acos(-1.0)
     31 //typedef
     32 typedef long long LL;
     33 typedef unsigned long long ULL;
     34 //const
     35 const int N=8010;
     36 const int INF=0x3f3f3f3f;
     37 const int MOD=5000,STA=100010;
     38 const LL LNF=1LL<<60;
     39 const double EPS=1e-8;
     40 const double OO=1e15;
     41 const int dx[4]={-1,0,1,0};
     42 const int dy[4]={0,1,0,-1};
     43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
     44 //Daily Use ...
     45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
     46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
     47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
     48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
     49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
     50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
     51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
     52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
     53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
     54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
     55 //End
     56 
     57 int first[N*2],next[20000*2],vis[N*2],S[N*2];
     58 int n,m,mt,cnt;
     59 
     60 struct Edge{
     61     int u,v;
     62 }e[20000*2];
     63 
     64 void adde(int a,int b)
     65 {
     66     e[mt].u=a,e[mt].v=b;
     67     next[mt]=first[a];first[a]=mt++;
     68 }
     69 
     70 int dfs(int u)
     71 {
     72     if(vis[u^1])return 0;
     73     if(vis[u])return 1;
     74     int i;
     75     vis[u]=1;
     76     S[cnt++]=u;
     77     for(i=first[u];i!=-1;i=next[i]){
     78         if(!dfs(e[i].v))return 0;
     79     }
     80     return 1;
     81 }
     82 
     83 int Twosat()
     84 {
     85     int i,j;
     86     for(i=0;i<n;i+=2){
     87         if(vis[i] || vis[i^1])continue;
     88         cnt=0;
     89         if(!dfs(i)){
     90             while(cnt)vis[S[--cnt]]=0;
     91             if(!dfs(i^1))return 0;
     92         }
     93     }
     94     return 1;
     95 }
     96 
     97 int main()
     98 {
     99  //   freopen("in.txt","r",stdin);
    100     int i,j,a,b;
    101     while(~scanf("%d%d",&n,&m))
    102     {
    103         n<<=1;
    104         mem(first,-1);mt=0;
    105         mem(vis,0);
    106         for(i=0;i<m;i++){
    107             scanf("%d%d",&a,&b);
    108             a--,b--;
    109             adde(a,b^1);
    110             adde(b,a^1);
    111         }
    112 
    113         if(Twosat()){
    114             for(i=0;i<n;i+=2){
    115                 printf("%d
    ",(vis[i]?i:i^1)+1);
    116             }
    117         }
    118         else puts("NIE");
    119     }
    120     return 0;
    121 }
  • 相关阅读:
    第十四章 构建自定义的同步工具(待续)
    第十三章 显示锁(待续)
    第十二章 并发程序的测试(待续)
    第十一章 性能与可伸缩性(待续)
    第十章 避免活跃性危险(待续)
    第九章 图形用户界面的并行化(待续)
    第八章 线程池的使用(待续)
    第七章 取消与关闭(待续)
    NOIp 2015真题模拟赛 By cellur925
    [USACO10MAR]伟大的奶牛聚集Great Cow Gat…【树形dp】By cellur925
  • 原文地址:https://www.cnblogs.com/zhsl/p/3175931.html
Copyright © 2020-2023  润新知