zoukankan      html  css  js  c++  java
  • HDU3666 差分约束

    THE MATRIX PROBLEM

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 8418    Accepted Submission(s): 2179


    Problem Description
    You have been given a matrix CN*M, each element E of CN*M is positive and no more than 1000, The problem is that if there exist N numbers a1, a2, … an and M numbers b1, b2, …, bm, which satisfies that each elements in row-i multiplied with ai and each elements in column-j divided by bj, after this operation every element in this matrix is between L and U, L indicates the lowerbound and U indicates the upperbound of these elements.
     
    Input
    There are several test cases. You should process to the end of file.
    Each case includes two parts, in part 1, there are four integers in one line, N,M,L,U, indicating the matrix has N rows and M columns, L is the lowerbound and U is the upperbound (1<=N、M<=400,1<=L<=U<=10000). In part 2, there are N lines, each line includes M integers, and they are the elements of the matrix.

     
    Output
    If there is a solution print "YES", else print "NO".
     
    Sample Input
    3 3 1 6
    2 3 4
    8 2 6
    5 2 9
     
    Sample Output
    YES
     
    Source
     题意:
    n*m的矩阵c,是否存在n个数a1,a2,...an,和m个数b1,b2,...bm使得L<=cij*ai/bj<=R.
    代码:
    //显然不满足差分约束的条件,可以L<=cij*ai/bj<=R两边除cij(cij>0)后取对数得到
    //log(L/cij)<=log(ai)-log(bj)<=log(R/cij).只求存不存在就行。但是本体如果用stl
    //的queue写spfa会超时(可以用节点出队次数小于sqrt(n)判断),可以自己定义一个栈来存储。
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    using namespace std;
    const int maxn=402;
    const double inf=100000008;
    int n,m,tol,head[maxn*2+10],cnt[maxn*2+10],stk[maxn*maxn];
    double L,R,dis[maxn*2+10];
    bool mark[maxn*2+10];
    struct node
    {
        int to,next;
        double val;
    }nodes[360010];
    void Add(int a,int b,double c)
    {
        nodes[tol].to=b;
        nodes[tol].val=c;
        nodes[tol].next=head[a];
        head[a]=tol++;
    }
    bool spfa(int s)
    {
        for(int i=0;i<=n+m;i++){
            dis[i]=inf;
            cnt[i]=0;
            mark[i]=0;
        }
        int top=0;
        stk[++top]=s;
        mark[s]=1;cnt[s]++;dis[s]=0;
        while(top>0){
            int u=stk[top--];
            mark[u]=0;
            for(int i=head[u];i!=-1;i=nodes[i].next){
                int v=nodes[i].to;
                if(dis[v]>dis[u]+nodes[i].val){
                    dis[v]=dis[u]+nodes[i].val;
                    if(!mark[v]){
                        mark[v]=1;
                        if(++cnt[v]>=n+m) return 0;
                        stk[++top]=v;
                    }
                }
            }
        }
        return 1;
    }
    int main()
    {
        while(scanf("%d%d%lf%lf",&n,&m,&L,&R)==4){
            tol=0;
            memset(head,-1,sizeof(head));
            double x;
            for(int i=1;i<=n;i++){
                for(int j=1;j<=m;j++){
                    scanf("%lf",&x);
                    Add(j+n,i,log(R/x));
                    Add(i,j+n,-log(L/x));
                }
            }
            for(int i=1;i<=n+m;i++)//加一个公共源点0
                Add(0,i,0);
            if(spfa(0)) printf("YES
    ");
            else printf("NO
    ");
        }
        return 0;
    }
  • 相关阅读:
    Nodejs 进阶:Express 常用中间件 body-parser 实现解析
    Nodejs进阶:express+session实现简易身份认证
    Node 进阶:express 默认日志组件 morgan 从入门使用到源码剖析
    Nodejs进阶:如何玩转子进程(child_process)
    express+session实现简易身份认证
    你真的了解UIViewController生命周期吗?
    你真的了解UIGestureRecognizer吗?
    你真的了解UIEvent、UITouch吗?
    你真的了解UIScrollView吗?
    你真的了解UITextView吗?
  • 原文地址:https://www.cnblogs.com/--ZHIYUAN/p/6536119.html
Copyright © 2011-2022 走看看