标签:read eps tchar ble scl queue www cti int
https://www.luogu.org/problemnew/show/P4171
意识到图中只有两种不同的菜系:满和汉
并且检查员类似于一个约束,可以发现这就是一个2-sat模型,满和汉分别对应true和false
由于只是检查可行性,只需要判断存在点的true个false存在同一个强连通分量即可。
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();} while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 210; const int maxm = 2010; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; struct Edge{ int to,next; }edge[maxm << 2]; int head[maxn],tot; int Low[maxn],dfn[maxn],Stack[maxn],Belong[maxn]; int Index,top,scc; bool Instack[maxn]; void init(){ for(int i = 0 ; i <= (N << 1) ; i ++){ head[i] = -1; Low[i] = dfn[i] = Belong[i] = Instack[i] = 0; } tot = scc = top = Index = 0; } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } 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); } } int main(){ // ios::sync_with_stdio(false); int T = read(); while(T--){ cin >> N >> M; init(); for(int i = 1; i <= M ; i ++){ char op1,op2; int x,y; cin >> op1 >> x >> op2 >> y; int flag1 = op1 == ‘m‘?1:0; int flag2 = op2 == ‘m‘?1:0; add(x + N * (flag1 ^ 1),y + N * (flag2 & 1)); add(y + N * (flag2 ^ 1),x + N * (flag1 & 1)); } for(int i = 1; i <= (N << 1) ; i ++) if(!dfn[i]) Tarjan(i); int flag = 1; for(int i = 1; i <= N && flag; i ++){ if(Belong[i] == Belong[i + N]) flag = 0; } if(flag) puts("GOOD"); else puts("BAD"); } return 0; }
标签:read eps tchar ble scl queue www cti int
原文地址:https://www.cnblogs.com/Hugh-Locke/p/10774101.html