学步园

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 ≤
i ≤ m) line of them are a_i-1 ,b_i-1 andc_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

题意：

读一段程序，问怎样取x数组的值使得输出的dep最大。

思路：

要使dep最大，则要尽量满足 x[a[dep]] + x[b[dep]] != c[dep] ，x只有两种取值，很快想到2-sat模型，二分答案，根据c的值建好图，看是否满足条件就够了。

代码：

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

int n,m,num,flag,ans;
int head[maxn];
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(head,0,sizeof(head));
    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;
    g[num].next = head[u];
    head[u] = num;
}
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)
        {
            addedge(a[i],b[i]+n);
            addedge(b[i],a[i]+n);
        }
        else if(c[i]==1)
        {
            addedge(a[i],b[i]);
            addedge(a[i]+n,b[i]+n);
            addedge(b[i],a[i]);
            addedge(b[i]+n,a[i]+n);
        }
        else
        {
            addedge(a[i]+n,b[i]);
            addedge(b[i]+n,a[i]);
        }
    }
    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;
}