## zoj 3422 Go Deeper （ 二分+2-sat ）

2014年08月29日 ⁄ 综合 ⁄ 共 2602字 ⁄ 字号 评论关闭

Go Deeper

Time Limit: 2 Seconds
Memory Limit:
65536 KB

Here is a procedure's pseudocode:

```	   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
```

In this code n is an integer. a, b, c and
x
are 4 arrays of integers. The index of array always starts from 0. Array
a
and b consist of non-negative integers smaller than n. Arrayx consists of only 0 and 1. Array
c consists of only 0, 1 and 2. The lengths of arraya, b and
c are m while the length of array x is n.

Given the elements of array a, b, and c, when we call the procedure go(0,n ,
m) what is the maximal possible value does the procedure output?

Input

There are multiple test cases. The first line of input is an integer T (0 <T ≤ 100), indicating the number of test cases. Then
T test cases follow. Each case starts with a line of 2 integersn and
m (0 < n ≤ 200, 0 < m ≤ 10000). Then m lines of 3 integers follow. Thei-th(1 ≤
im) line of them are ai-1 ,bi-1 andci-1 (0 ≤
ai-1, bi-1 <n, 0 ≤ ci-1 ≤ 2).

Output

For each test case, output the result in a single line.

Sample Input

```3
2 1
0 1 0
2 1
0 0 0
2 2
0 1 0
1 1 2

```

Sample Output

```1
1
2
```

```#include <cstdio>
#include <cstring>
#define maxn 405
#define MAXN 200005
using namespace std;

int n,m,num,flag,ans;
int scc[maxn];
int vis[maxn];
int stack1[maxn];
int stack2[maxn];
int a[MAXN],b[MAXN],c[MAXN];
struct edge
{
int v,next;
} g[MAXN];

void init()
{
memset(vis,0,sizeof(vis));
memset(scc,0,sizeof(scc));
stack1[0] = stack2[0] = num = 0;
flag = 1;
}
void addedge(int u,int v)
{
num++;
g[num].v = v;
}
void dfs(int cur,int &sig,int &cnt)
{
if(!flag) 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].v]) dfs(g[i].v,sig,cnt);
else
{
if(!scc[g[i].v])
{
while(vis[stack2[stack2[0]]] > vis[g[i].v])
stack2[0] --;
}
}
}
if(stack2[stack2[0]] == cur)
{
stack2[0] --;
++cnt;
do
{
scc[stack1[stack1[0]]] = cnt;
int tmp = stack1[stack1[0]];
if((tmp >= n && scc[tmp - n] == cnt) || (tmp < n && scc[tmp + n] == cnt))
{
flag = false;
return;
}
}
while(stack1[stack1[0] --] != cur);
}
}
void Twosat()
{
int i,sig,cnt;
sig = cnt = 0;
for(i=0; i<n+n&&flag; i++)
{
if(!vis[i]) dfs(i,sig,cnt);
}
}
bool isok(int mid)
{
int i,j,t;
init();
num=0;
for(i=0; i<=mid; i++)
{
if(c[i]==0)
{
}
else if(c[i]==1)
{
}
else
{
}
}
Twosat();
if(flag) return true ;
return false ;
}
void solve()
{
int i,j,t,le,ri,mid;
le=0;
ri=m-1;
while(le<ri)
{
mid=(le+ri+1)>>1;
if(isok(mid)) le=mid;
else ri=mid-1;
}
ans=le+1;
}
int main()
{
int i,j,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();
printf("%d\n",ans);
}
return 0;
}
```