[2-sat][裸题] HDU3062 Party

这题只判断可行性，不用输出解，应该是2-sat入门题里最水的一道....

代码：

//n个元素。i=0~n-1.
//i表示0、i+n表示1、i+2n表示0'、i+3n表示1'。

//1wa。这题竟然有多组数据。...

#include
#include
#include
using namespace std;

#define MAXN 4002	//4倍的点
//#define MAXM 8000002		//n*n/2
#define MAXM 30000100		//???

struct node{ int u, v; }a[MAXM];
int first[MAXN], next[MAXM], idx;		//idx 边下标
int n;		//n对夫妻
int m;	//m对矛盾关系

void addedge(int u, int v)
{
	a[idx].u = u, a[idx].v = v;
	next[idx] = first[u];
	first[u] = idx++;
}

int dfn[MAXN], low[MAXN];
int stack[MAXN], ins[MAXN];
int belong[MAXN], cnt;		//属于什么连通分量
int top, num;
void tarjin(int u)
{
	dfn[u] = low[u] = ++num;
	ins[u] = 1;
	stack[top++] = u;
	for(int e=first[u]; e!=-1; e=next[e])
	{
		int v = a[e].v;
		if(!dfn[v])
		{
			tarjin(v);
			low[u] = min(low[u], low[v]);
		}
		else if(ins[v])
		{
			low[u] = min(low[u], dfn[v]);
		}
	}
	if(dfn[u] == low[u])
	{
		while(1)
		{
			int v = stack[--top];
			ins[v] = 0;
			belong[v] = cnt;			//标记分量belong
			if(u == v) break;
		}
		cnt++;
	}
}

int solve()
{
	for(int i=0; i<4*n; i++)	//4*n 总点数
	{
		if(!dfn[i])
			tarjin(i);
	}
	for(int i=0; i<2*n; i++)	//遍历 女、男
	{
		if(belong[i] == belong[i+2*n])
		{
			return 0;
		}
	}
	return 1;
}

int read()
{
	memset(belong, 0, sizeof(belong));
	memset(dfn, 0, sizeof(dfn));
	memset(first, -1, sizeof(first));
	memset(ins, 0, sizeof(ins));
	num = cnt = 0;
	top = 0;

	if(scanf("%d%d", &n, &m)!=EOF)
	{
	idx = 0;		//初始化边编号
	for(int i=0; i男
		addedge(a1, a2+2*n);
		addedge(a2, a1+2*n);
	}
	return 1;
	}
	else return 0;
}
int main()
{
	while(read())
	{
	if(solve()) printf("YES\n");
	else printf("NO\n");
	}
}

转载于:https://www.cnblogs.com/tclh123/archive/2011/10/28/2587053.html