vector邻接表: ; struct Edge{ int u,v,w; Edge(int _u=0,int _v=0,int _w=0){u=_u,v=_v,w=_w;} }; vector<Edge> E; vector<int> G[maxn]; void init(int l,int r) { E.clear(); for(int i=l;i<=r;i++) G[i].clear(); } void addedge(int u,int v,int w) { E.pus…

 Swap Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2819 Description Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to ma…

Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to make all the diagonal entries equal to 1? InputThere are several test cases in the input. The first line of each test case is an…

题目链接 畅通工程,可以用dijkstra算法实现. 听说spfa很好用,来水一发 邻接矩阵实现: #include <stdio.h> #include <algorithm> #include <cmath> #include <cstring> #include <deque> #include <iomanip> #include <iostream> #include <list> #include…

Pet Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1809    Accepted Submission(s): 874 Problem Description One day, Lin Ji wake up in the morning and found that his pethamster escaped. He sear…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2819 题目大意:给你一个n*n的01矩阵,问是否可以通过任意交换整行或者整列使得正对角线上都是1. 解题思路: 按行列建图,求最大匹配,若最大匹配数<n,则肯定无解(矩阵的秩要等于n才有解).若有解,则肯定可以通过只交换行或者交换列得到.所以我们从i=1 to n检查link[i]=i(link[i]表示第i列的1对应的行号)是否成立,否则找到link[j]=i交换link[i]和link[j]并…

题目链接:https://cn.vjudge.net/problem/HDU-2819 Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to make all the diagonal entries equal to 1? InputThere are several test cases in the in…

Swap http://acm.hdu.edu.cn/showproblem.php?pid=2819 Special Judge Problem Description Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to make all the diagonal entries equal to 1?  …

题意:一个N*N的01矩阵,行与行.列与列之间可以互换.要求变换出一个对角线元素全为1的矩阵,给出互换的行号或列号. 分析:首先一个矩阵若能构成对角线元素全为1,那么矩阵的秩为N,秩小于N的情况无解.所以一个矩阵可以仅通过行变换不能得到最后结果,那么仅通过列变换或者行列变换都不能得到. 可以将行号看作二分图的X部,列号对应二分图Y部,那么矩阵Mij为1就可以视作第i行可以与第j列相匹配.而构成对角线全1的情况就是行与列之间有N对匹配关系,若小于N则无解. 根据所给的矩阵建二分图跑匈牙利,然后可以…

 棋盘游戏 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1281 Description 小希和Gardon在玩一个游戏:对一个N*M的棋盘,在格子里放尽量多的一些国际象棋里面的“车”,并且使得他们不能互相攻击,这当然很简单,但是Gardon限制了只有某些格子才可以放,小希还是很轻松的解决了这个问题(见下图)注意不能放车的地方不影响车…