出题人真是脑洞堪比黑洞 (然后自己也被吸进去了 理解一遍题意 三个条件可以转化为 1的出度是1, n的入度是1, 2~n-1的出度等于入度 不难发现1-n的最短路符合题意 然而其实还有另一种情况 1为起止点的最短闭环+n为起止点的最短闭环同样满足要求 因此取两者的min作为结果 #include <iostream> #include <string> #include <cstdio> #include <cmath> #include <cstri…

0 or 1 题目链接: Rhttp://acm.hust.edu.cn/vjudge/contest/122685#problem/R Description Given a n*n matrix C ij (1<=i,j<=n),We want to find a n*n matrix X ij (1<=i,j<=n),which is 0 or 1. Besides,X ij meets the following conditions: 1.X 12+X 13+...X 1…

HDU - 4370 参考:https://www.cnblogs.com/hollowstory/p/5670128.html 题意: 给定一个矩阵C, 构造一个A矩阵,满足条件: 1.X12+X13+...X1n=1 2.X1n+X2n+...Xn-1n=1 3.for each i (1<i<n), satisfies ∑Xki (1<=k<=n)=∑Xij (1<=j<=n). 使得∑Cij*Xij(1<=i,j<=n)最小. 思路: 理解条件之前先…

0 or 1 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2811    Accepted Submission(s): 914 Problem Description Given a n*n matrix Cij (1<=i,j<=n),We want to find a n*n matrix Xij (1<=i,j<…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4370 题目大意:有一个n*n的矩阵Cij(1<=i,j<=n),要找到矩阵Xij(i<=1,j<=n)满足以下条件: 1.X 12+X 13+...X 1n=1  2.X 1n+X 2n+...X n-1n=1  3.for each i (1<i<n), satisfies ∑X ki (1<=k<=n)=∑X ij (1<=j<=n). 举个例子…

Description Given a n*n matrix C ij (1<=i,j<=n),We want to find a n*n matrix X ij (1<=i,j<=n),which is 0 or 1. Besides,X ij meets the following conditions: 1.X 12+X 13+...X 1n=1 2.X 1n+X 2n+...X n-1n=1 3.for each i (1<i<n), satisfies ∑X…

[题目链接](http://acm.hdu.edu.cn/showproblem.ph Problem Description Given a n/n matrix Cij (1<=i,j<=n),We want to find a n/n matrix Xij (1<=i,j<=n),which is 0 or 1. Besides,Xij meets the following conditions: 1.X12+X13+...X1n=1 2.X1n+X2n+...Xn-1n=…

题意:给一个N*N的矩阵C,和一个N*N的只由0和1组成的矩阵X. X满足以下条件: 1.X 12+X 13+...X 1n=1  2.X 1n+X 2n+...X n-1n=1  3.任意 i (1<i<n), 都有 ∑X ki (1<=k<=n)=∑X ij (1<=j<=n). 求∑C ij*X ij(1<=i,j<=n) 的最小值(1<n<=300). 分析:乍一看也想不到这是道最短路的题.但将模型转换后,则易想到方法.根据01矩阵X满足…

题目传送门 题意:题目巨晦涩的传递出1点和n点的初度等于入度等于1, 其余点出度和入度相等 分析:求最小和可以转换成求最短路,这样符合条件,但是还有一种情况.1点形成一个环,n点也形成一个环,这样也是可以的,这样SPFA要稍微修改点,d[s] = INF,表示可以更新. #include <bits/stdc++.h> using namespace std; const int N = 3e2 + 5; const int INF = 0x3f3f3f3f; int w[N][N]; int…

很有挑战的一题 直接暴力建图的话毫无疑问O(n^2)会TLE 每层虚拟一个点又会让没有点的层也能连过去 参考kuangbin菊苣的方法每层用了两个虚拟点 n+i*2-1 是入口 n+i*2 是出口 然后建单向边就可以了 VA了一次 因为MAXN应该比数据量大两倍 不小心忽略了 至于MAXM直接开到了1e7 #include <iostream> #include <string> #include <cstdio> #include <cmath> #inc…