Light OJ 1251 Forming the Council 2-SAT输出任意一组解

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题目来源：Light OJ 1251 Forming the Council

题意：若干了条件至少满足一个 求是否有方案 输出任意一种可能的方案 留下的人的个数

思路：2-SAT基础题

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 100010;
int n, m;
vector <int> G[maxn*2];
bool mark[maxn*2];
int S[maxn*2], c;
int a[maxn], b[maxn], sum;
int ans[maxn];
bool dfs(int x)
{
	if(mark[x^1])
		return false;
	if(mark[x])
		return true;
	mark[x] = true;
	S[c++] = x;
	for(int i = 0; i < G[x].size(); i++)
		if(!dfs(G[x][i]))
			return false;
	return true;
}

void init()
{
	for(int i = 0; i < n*2; i++)
		G[i].clear();
	memset(mark, 0, sizeof(mark));
}

void AddEdge(int u, int v, int x, int y)
{
	u = u * 2 + x;
	v = v * 2 + y;
	G[u].push_back(v^1);
	G[v].push_back(u^1);
}
bool solve()
{
	for(int i = 0; i < n*2; i += 2)
	{
		if(!mark[i] && !mark[i+1])
		{
			c = 0;
			if(!dfs(i))
			{
				while(c > 0)
					mark[S[--c]] = false;
				if(!dfs(i+1))
					return false;
			}
		}
	}
	return true;
}

int get(char *s)
{
	int ans = 0;
	for(int i = 1; s[i]; i++)
		ans = ans * 10 + s[i] - '0';
	return ans;
}
int main()
{
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d %d", &m, &n);
		init();
		while(m--)
		{
			char s1[100], s2[100];
			scanf("%s %s", s1, s2);
			int u = get(s1);
			int v = get(s2);
			u--;
			v--;
			int x, y;
			if(s1[0] == '+')
				x = 1;
			else
				x = 0;
			if(s2[0] == '+')
				y = 1;
			else
				y = 0;
			AddEdge(u, v, x^1, y^1);
		}
		if(!solve())
		{
			printf("Case %d: No\n", cas++);
			continue;
		}
		int l = 0;
		for(int i = 0; i < n; i++)
			if(mark[i*2+1])
				ans[l++] = i;
		printf("Case %d: Yes\n", cas++);
		printf("%d", l);
		for(int i = 0; i < l; i++)
			printf(" %d", ans[i]+1);
		puts("");
	}
	return 0;
}

Light OJ 1251 Forming the Council 2-SAT输出任意一组解,布布扣,bubuko.com

Light OJ 1251 Forming the Council 2-SAT输出任意一组解

标签：c   class   blog   code   a   http   

原文地址：http://blog.csdn.net/u011686226/article/details/28909767