HDU 4115(Eliminate the Conflict-石头剪刀布,经典2sat问题)

有向图双联通性与逻辑判定 - 石头剪刀布

• 玩剪刀石头布，n把

• 给了A这n把都出了什么，问你B能否会赢

• 其中A会限制B某些局数出的要相同

• 某些局数出的要不同

• 只要B满足他的限制，并且没没有输掉任何一把就算赢

A每局有2个状态，赢和平

令xi表示第i局A是否赢

2sat分8种情况讨论

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase) printf("Case #%d: ",kcase);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
ll Abs(ll x){return (x<0)?-x:x;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}

#define MAXN (100000+10)
struct TwoSat{
int n;
vi G[MAXN*2];
bool mark[MAXN*2];
int S[MAXN*2],c;
bool dfs(int x) {
if (mark[x^1]) return 0;
if (mark[x]) return 1;
mark[x]=1;
S[c++]=x;
int sz=SI(G[x]);
Rep(i,sz) if (!dfs(G[x][i])) return 0;
return 1;
}
void init(int n) {
this -> n = n;
Rep(i,n*2) G[i].clear();
MEM(mark)
}
//x = xval or y = yval
void add_clause(int x,int xval,int y,int yval) {
x=x*2+xval;
y=y*2+yval;
G[x^1].pb(y);
G[y^1].pb(x);
}
bool solve() {
for(int i=0;i<n*2;i+=2) {
if (!mark[i] && !mark[i+1]) {
c=0;
if (!dfs(i)) {
while (c) mark[S[--c]] = 0;
if (!dfs(i+1)) return 0;
}
}
}
return 1;
}
}solver;
int n,m;
int a[MAXN];
// win,los
int h[4]={0,2,3,1};
int main()
{
//  freopen("C.in","r",stdin);
//  freopen(".out","w",stdout);
int T=read();
For(kcase,T) {
n=read(),m=read();
solver.init(3*n);
For(i,n) {
a[i]=read();
}
For(i,m) {
int u=read(),v=read(),k=read();
int wu=h[a[u]],lu=a[u],wv=h[a[v]],lv=a[v];
if (!k) {
if (wu!=wv) solver.add_clause(u,0,v,0);
if (wu!=lv) solver.add_clause(u,0,v,1);
if (lu!=wv) solver.add_clause(u,1,v,0);
if (lu!=lv) solver.add_clause(u,1,v,1);
} else {
if (wu==wv) solver.add_clause(u,0,v,0);
if (wu==lv) solver.add_clause(u,0,v,1);
if (lu==wv) solver.add_clause(u,1,v,0);
if (lu==lv) solver.add_clause(u,1,v,1);
}
}

Pr(kcase)
if (solver.solve()) puts("yes"); else puts("no");
}
return 0;
}