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;
}