zoj3422Go Deeper(2-sat + 二分)

题目请戳这里

题目大意:

go(int dep, int n, int m)  
	   begin  
	      output the value of dep. 
	      if dep < m and x[a[dep]] + x[b[dep]] != c[dep] then go(dep + 1, n, m)
	   end 

读上面程序段,yy出函数功能。数组a,b,c长度为m,x长度为n。数组a,b中元素范围[0,n - 1],数组c元素为0或1或2。x数组元素为1或0。求能输出的最大的m。

 

题目分析:2-sat,还是比较裸的吧。要求最大的m,所以对长度m二分。

详情请见代码:

 

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 405;
const int M = 10005;

struct node
{
    int to,next;
}g[M];
int head[N],stack1[N],stack2[N],vis[N],scc[N];
int n,m,num;
bool flag;
int a[M],b[M],c[M];
void init()
{
    memset(head,-1,sizeof(head));
    flag = true;
    memset(vis,0,sizeof(vis));
    memset(scc,0,sizeof(scc));
    stack1[0] = stack2[0] = 0;
    num = 0;
}
void build(int s,int e)
{
    g[num].to = e;
    g[num].next = head[s];
    head[s] = num ++;
}
void dfs(int cur,int &sig,int &cnt)
{
    if(flag == false)
        return;
    vis[cur] = ++ sig;
    stack1[++stack1[0]] = cur;
    stack2[++stack2[0]] = cur;
    for(int i = head[cur];~i;i = g[i].next)
    {
        if(!vis[g[i].to])
            dfs(g[i].to,sig,cnt);
        else
        {
            if(scc[g[i].to] == 0)
                while(vis[stack2[stack2[0]]] > vis[g[i].to])
                    stack2[0] --;
        }
    }
    if(stack2[stack2[0]] == cur)
    {
        stack2[0] --;
        cnt ++;
        do
        {
            scc[stack1[stack1[0]]] = cnt;
            if(scc[stack1[stack1[0]]^1] == cnt)
            {
                flag = false;
                return;
            }
        }while(stack1[stack1[0] --] != cur);
    }
}
void Gabow()
{
    int i,sig,cnt;
    sig = cnt = 0;
    for(i = 0;i < n + n && flag;i ++)
        if(!vis[i])
            dfs(i,sig,cnt);
}
void solve()
{
    int l,r,mid;
    int ans,i;
    l = 0;r = m;
    while(l <= r)
    {
        mid = (l + r)>>1;
        init();
        for(i = 0;i < mid;i ++)
        {
            int u = a[i]<<1;
            int v = b[i]<<1;
            if(c[i] == 0)// !=0
            {
                build(u,v^1);
                build(v,u^1);
            }
            if(c[i] == 1)// != 1
            {
                build(u,v);
                build(v,u);
                build(u^1,v^1);
                build(v^1,u^1);
            }
            if(c[i] == 2)// != 2
            {
                build(u^1,v);
                build(v^1,u);
            }
        }
        Gabow();
        if(flag)
        {
            ans = mid;
            l = mid + 1;
        }
        else
            r = mid - 1;
    }
    printf("%d\n",ans);
}
int main()
{
    int i,t;
    scanf("%d",&t);
    while(t --)
    {
        scanf("%d%d",&n,&m);
        for(i = 0;i < m;i ++)
            scanf("%d%d%d",&a[i],&b[i],&c[i]);
        solve();
    }
    return 0;
}


 

 

posted on 2013-09-30 22:06  you Richer  阅读(241)  评论(0编辑  收藏  举报