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  • Go Deeper HDU

    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 ai-1 ,bi-1 and ci-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
     
    Author
    CAO, Peng
     
    Source
     
    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
    ", a);
    #define plld(a) printf("%lld
    ", a);
    #define pc(a) printf("%c
    ", a);
    #define ps(a) printf("%s
    ", 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 ai-1 ,bi-1 and ci-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
     
    Author
    CAO, Peng
     
    Source
     
    Recommend
    zhouzeyong
     
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  • 原文地址:https://www.cnblogs.com/WTSRUVF/p/9796628.html
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