Go Deeper（2010成都现场赛题）（2-sat）

G - Go Deeper

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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 mwhile 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 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

题目大意：
　
给定一些方程，x[]数组未知，求前面最多能够有多少方程x[a]+x[b]！=c能够被满足。
　其中 c=0,1,2 
　　x[]={0,1}
相当于裸的2sat问题，加上二分
　　
　　强烈建议阅读 kuangbin大神对2-sat的总结：http://www.cnblogs.com/kuangbin/archive/2012/10/05/2712429.html

总的来说，就是当 a or b 时 连接 a' -> b 与 b' -> a 的边，然后进行强连通判断是否出现 a 与 a' ...在同一个连通分量中，若在则不可能。

建立数a的两个状态，即a与a'，相当于x[a]=1 和 x[a']=0

x[a]+x[b]!=0   =>   a or b   =>   a'->b 且 a->b'
x[a]+x[b]!=1   =>   (a and b) or (a' and b') == a or b' 且 a' or b   =>   a'->b' 且 b->a 且 a->b 且 b'->a
x[a]+x[b]!=2   =>   a' or b'   =>   a->b' 且 a'->b

按上面来建图判断即可

#include<cstdio>
#include<cstring>
int e[50000],pd[50000],be[800],ne[50000],all;
int dfn[800],low[800],instack[800],belong[800],stack[800],stak,curr,num;
int a,b,c,n,m,l,r,mid,flag;
void add(int x,int y,int p){
    e[++all]=y;
    pd[all]=p;
    ne[all]=be[x];
    be[x]=all;
}
void tarjan(int x){
    instack[x]=1;
    stack[++stak]=x;
    dfn[x]=low[x]=++curr;
    for(int j=be[x];j!=0;j=ne[j])
    if(pd[j]<=mid){
        if(!dfn[e[j]]){
            tarjan(e[j]);
            if(low[x]>low[e[j]]) low[x]=low[e[j]];
        }else if(instack[e[j]]&&low[x]>low[e[j]])
            low[x]=low[e[j]];
    }
    if(dfn[x]==low[x]){
        int j;
        ++num;
        do{
            j=stack[stak--];
            instack[j]=0;
            belong[j]=num;
        }while(j!=x);
    }
}
int solve(){
    curr=stak=num=0;
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(instack,0,sizeof(instack));
    for(int i=0;i<2*n;i++)
        if(!dfn[i]) tarjan(i);
    flag=0;
    for(int i=0;i<n;i++)
        if(belong[2*i]==belong[2*i+1]){
            flag=1;
            break;
        }
    return flag;
}
int main()
{
    int tt;
    scanf("%d",&tt);
    while(tt--){
        scanf("%d%d",&n,&m);
        all=0;
        memset(e,0,sizeof(e));
        memset(be,0,sizeof(be));
        memset(ne,0,sizeof(ne));
        memset(pd,0,sizeof(pd));
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&a,&b,&c);
            switch (c){
            case 0: add(2*a+1,2*b,i);
                    add(2*b+1,2*a,i);
                    break;
            case 1: add(2*a,2*b,i);
                    add(2*b+1,2*a+1,i);
                    add(2*b,2*a,i);
                    add(2*a+1,2*b+1,i);
                    break;
            case 2: add(2*a,2*b+1,i);
                    add(2*b,2*a+1,i);
                    break;
            }
        }
        l=0;r=m-1;
        while(l<r-1){
            mid=(l+r)/2;
            if(solve()) r=mid; else l=mid;
        }
        mid=r;
        if(!solve()) printf("%d
",r+1);
        else printf("%d
",l+1);
    }
    return 0;
}