Go Deeper HDU - 3715（2 - sat 水题 妈的 智障）

Go Deeper

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3435    Accepted Submission(s): 1125

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

 

 

Recommend

zhouzeyong

 

 

 

解析：

　　一定要明确 是哪两个点

　　然后建图一定要明确怎么建

　　二分定要写对

 

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define pd(a) printf("%d\n", a);
#define plld(a) printf("%lld\n", a);
#define pc(a) printf("%c\n", a);
#define ps(a) printf("%s\n", a);
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1e5 + 10, INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
int n, m;
int a[maxn], b[maxn], c[maxn];
vector<int> G[maxn];
int sccno[maxn], low[maxn], vis[maxn], scc_clock, scc_cnt;
stack<int> S;
void init()
{
    for(int i = 0; i < maxn; i++) G[i].clear();
    mem(sccno, 0);
    mem(low, 0);
    mem(vis, 0);
    scc_clock = scc_cnt = 0;
}

void dfs(int u)
{
    low[u] = vis[u] = ++scc_clock;
    S.push(u);
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(!vis[v])
        {
            dfs(v);
            low[u] = min(low[u], low[v]);
        }
        else if(!sccno[v])
            low[u] = min(low[u], vis[v]);
    }
    if(vis[u] == low[u])
    {
        scc_cnt++;
        for(;;)
        {
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            if(x == u) break;
        }
    }
}

void build(int mid)
{
    for(int i = 0; i <= mid; i++)
    {
        if(c[i] == 2)
        {
            G[a[i] << 1 | 1].push_back(b[i] << 1);
            G[b[i] << 1 | 1].push_back(a[i] << 1);
        }
        else if(c[i] == 1)
        {
            G[a[i] << 1 | 1].push_back(b[i] << 1 | 1);
            G[b[i] << 1 | 1].push_back(a[i] << 1 | 1);
            G[a[i] << 1].push_back(b[i] << 1);
            G[b[i] << 1].push_back(a[i] << 1);
        }
        else if(c[i] == 0)
        {
            G[a[i] << 1].push_back(b[i] << 1 | 1);
            G[b[i] << 1].push_back(a[i] << 1 | 1);
        }
    }
}

bool check()
{
    for(int i = 0; i < n * 2; i += 2)
        if(sccno[i] == sccno[i + 1])
            return false;
    return true;
}

int main()
{
    int T;
    rd(T);
    while(T--)
    {
        init();
        rd(n), rd(m);
        for(int i = 0; i < m; i++)
        {
            rd(a[i]), rd(b[i]), rd(c[i]);
        }
        int l = 0, r = m;
        while(l + 1 < r)
        {
            init();
            int mid = (l + r) / 2;
            build(mid);
            for(int i = 0; i < n * 2; i++)
                if(!vis[i]) dfs(i);
            if(check()) l = mid;
            else r = mid;
        }
        pd(l + 1);

    }

    return 0;
}

 

 

 

 

Go Deeper

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3435    Accepted Submission(s): 1125

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

 

 

Recommend

zhouzeyong