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  • ACM HDU 3715 Go Deeper

    Go Deeper

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 309    Accepted Submission(s): 122


    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
     

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    zhouzeyong
     
     
    #include<stdio.h>
    #include
    <string.h>
    #define MAXN 405
    #define MAXM 160010
    struct Node
    {
    int from,to,next;
    }edge1[MAXM],edge2[MAXM];
    struct Node1
    {
    int a,b,c;
    }s[MAXM];
    int n,m,head1[MAXN],head2[MAXN],visit1[MAXN],visit2[MAXN],tol1,tol2;
    int Tcnt,Bcnt,Belong[MAXN],T[MAXN];
    void add(int a,int b)
    {
    edge1[tol1].from
    =a;edge1[tol1].to=b;edge1[tol1].next=head1[a];head1[a]=tol1++;
    edge2[tol2].from
    =b;edge2[tol2].to=a;edge2[tol2].next=head2[b];head2[b]=tol2++;
    }
    void dfs1(int i)
    {
    int j,u;
    visit1[i]
    =1;
    for(j=head1[i];j!=-1;j=edge1[j].next)
    {
    u
    =edge1[j].to;
    if(!visit1[u]) dfs1(u);
    }
    T[Tcnt
    ++]=i;
    }
    void dfs2(int i)
    {
    int j,u;
    visit2[i]
    =1;
    Belong[i]
    =Bcnt;
    for(j=head2[i];j!=-1;j=edge2[j].next)
    {
    u
    =edge2[j].to;
    if(!visit2[u]) dfs2(u);
    }
    }
    int main()
    {
    int i,ans,right,left,mid,ncase;
    scanf(
    "%d",&ncase);
    while(ncase--)
    {
    scanf(
    "%d%d",&n,&m);
    for(i=0;i<m;i++)
    scanf(
    "%d%d%d",&s[i].a,&s[i].b,&s[i].c);
    left
    =0;
    right
    =m;
    while(left<=right)
    {
    mid
    =(left+right)/2;
    for(i=0;i<2*n;i++)
    {
    head1[i]
    =-1;
    head2[i]
    =-1;
    visit1[i]
    =0;
    visit2[i]
    =0;
    }
    tol1
    =tol2=0;
    Tcnt
    =Bcnt=0;
    for(i=0;i<mid;i++)
    {
    if(s[i].c==0)
    {
    add(s[i].a,s[i].b
    +n);
    add(s[i].b,s[i].a
    +n);
    }
    else if(s[i].c==1)
    {
    add(s[i].a,s[i].b);
    add(s[i].b,s[i].a);
    add(s[i].a
    +n,s[i].b+n);
    add(s[i].b
    +n,s[i].a+n);
    }
    else if(s[i].c==2)
    {
    add(s[i].a
    +n,s[i].b);
    add(s[i].b
    +n,s[i].a);
    }
    }
    for(i=0;i<2*n;i++)
    if(!visit1[i]) dfs1(i);
    for(i=Tcnt-1;i>=0;i--)
    {
    if(!visit2[T[i]])
    {
    dfs2(T[i]);
    Bcnt
    ++;
    }
    }
    for(i=0;i<n;i++)
    {
    if(Belong[i]==Belong[i+n]) break;
    }
    if(i==n)
    {
    ans
    =mid;left=mid+1;
    }
    else right=mid-1;
    }
    printf(
    "%d\n",ans);
    }
    return 0;
    }

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  • 原文地址:https://www.cnblogs.com/kuangbin/p/2150058.html
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