HDU 3715 Go Deeper

Go Deeper

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2198    Accepted Submission(s): 722

Problem Description

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. Array x consists of only 0 and 1. Array c consists of only 0, 1 and 2. The lengths of array a, 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 the procedure may 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 integers n and m (0 < n ≤ 200, 0 < m ≤ 10000). Then m lines of 3 integers follow. The i-th(1 ≤ i ≤ m) line of them are a_i-1 ,b_i-1 and c_i-1 (0 ≤ a_i-1, b_i-1 < n, 0 ≤ c_i-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

 

Author

CAO, Peng

 

Source

2010 Asia Chengdu Regional Contest

 

题目的意思是求出符合条件的能去到的最大深度 。

然后 x 只有 0 ， 1 两种 。

明显就是一个   two - sat .

然后二分一个深度 ， 重新构图 ， 看下所有限制能否都符合 。

卡了一下二分 。 。 

构图我是反向的 ， 即 u - v 表示这两个条件不能共存 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>

using namespace std;
typedef long long LL;
const int N = 20010;
const int M = 800010;

int eh[N] , et[M], nxt[M] , tot ;
bool mark[N] ;
int n ,  s  , a[N] , b[N] ,c[N];
int st[N] , top  ;

void addedge(int u , int v )
{
    et[tot] = v , nxt[tot] = eh[u] , eh[u] = tot++;
    et[tot] = u , nxt[tot] = eh[v] , eh[v] = tot++;
}
void init()
{
    memset (mark , false , sizeof mark );
    tot = 0 ;
    memset ( eh  , -1 , sizeof eh );
}
// ----------------

bool dfs( int u )
{
    if( mark[u] ) return true ;
    if( mark[u^1] ) return false ;
    mark[u] = true;
    st[top++] = u ;
    for( int i = eh[u] ; ~i ; i = nxt[i] ){
        int v = et[i] ;
        if( !dfs( v^1 ) ) return false;
    }
    return true ;
}

bool solve( int m )
{
    init();
    for( int i = 0 ; i < m ; ++i ){
       if( c[i] == 0 ){
            addedge( 2*a[i] , 2*b[i] );
        }
        else if( c[i] == 1 ){
            addedge( (2*a[i])^1 , 2*b[i] );
            addedge( (2*b[i])^1 , 2*a[i] );
        }
        else {
            addedge( (2*a[i])^1 , (2*b[i])^1 );
        }
    }
    for( int i = 0 ; i < 2 * n ; i+=2 ){
        if( !mark[i] && !mark[i+1] ){
            top = 0 ;
            if( !dfs(i) ){
                while( top > 0 ) mark[ st[--top] ] = false ;
                if( !dfs(i+1) ) return false ;
            }
        }
    }
    return true;
}
void run()
{
    int  m ;
    scanf("%d%d",&n,&m);
    for( int i = 0 ; i < m ;++i ) scanf("%d%d%d",&a[i],&b[i],&c[i]);
    int  ans = 1 , l = 1 ,r = m ;

    while( l <= r ){
        int mid = ( l + r ) / 2;
//        cout << mid <<' '<<endl ;
        if( solve(mid) )
            l = mid + 1 , ans = mid ;
        else
            r = mid - 1;

    }
    printf("%d
",ans);
}


int main()
{
    #ifdef LOCAL
        freopen("in","r",stdin);
    #endif
    int _ ;
    scanf("%d",&_);
    while( _ -- )run();
    return 0;
}