Special Fish

Problem Description
There is a kind of special fish in the East Lake where is closed to campus of Wuhan University. It’s hard to say which gender of those fish are, because every fish believes itself as a male, and it may attack one of some other fish who is believed to be female
by it.

A fish will spawn after it has been attacked. Each fish can attack one other fish and can only be attacked once. No matter a fish is attacked or not, it can still try to attack another fish which is believed to be female by it.

There is a value we assigned to each fish and the spawns that two fish spawned also have a value which can be calculated by XOR operator through the value of its parents.

We want to know the maximum possibility of the sum of the spawns.
 
Input
The input consists of multiply test cases. The first line of each test case contains an integer n (0 < n <= 100), which is the number of the fish. The next line consists of n integers, indicating the value (0 < value <= 100) of each fish. The next n lines,
each line contains n integers, represent a 01 matrix. The i-th fish believes the j-th fish is female if and only if the value in row i and column j if 1.

The last test case is followed by a zero, which means the end of the input.
 
Output
Output the value for each test in a single line.
 
Sample Input
3
1 2 3
011
101
110
0
 
Sample Output
6
 
Author
momodi@whu
 
Source
 
Recommend

解题思路:

题意为有n条特殊的鱼,每一个鱼都有一个价值,假设鱼i ”觉得“ 鱼j 性别不同,那么就攻击它。生殖的后代的价值为 v[i] ^ v[j], 每条鱼仅仅能攻击或被攻击一次,问最后生殖的后代的最大价值为多少。

也是比較裸的二分图最大权不匹配,边i,j的权值等于 v[i] ^ v[j] 。

http://blog.csdn.net/sr_19930829/article/details/40650359

代码:

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int maxn=102;
const int inf=0x3f3f3f;
int nx,ny;//左右两边的点数
int v[maxn];//每条鱼的value
int g[maxn][maxn];//邻接矩阵
int linked[maxn];//右边的点和左边哪个点连接
int lx[maxn],ly[maxn];//左右点的标号
int slack[maxn];//slack[j]表示右边的点j的全部不在导出子图的边相应的lx[i]+ly[j]-w[i][j]的最小值
bool visx[maxn],visy[maxn]; bool DFS(int x)//hungary求增广路
{
visx[x]=true;
for(int y=0;y<ny;y++)
{
if(visy[y])
continue;
int tmp=lx[x]+ly[y]-g[x][y];
if(tmp==0)
{
visy[y]=true;
if(linked[y]==-1||DFS(linked[y]))
{
linked[y]=x;
return true;
}
}
else if(slack[y]>tmp)
slack[y]=tmp;
}
return false;
} int KM()
{
memset(linked,-1,sizeof(linked));
memset(ly,0,sizeof(ly));
for(int i=0;i<nx;i++)
{
lx[i]=-inf;
for(int j=0;j<ny;j++)
if(g[i][j]>lx[i])
lx[i]=g[i][j];
}
for(int x=0;x<nx;x++)
{
for(int y=0;y<ny;y++)
slack[y]=inf;
while(true)
{
memset(visx,0,sizeof(visx));
memset(visy,0,sizeof(visy));
if(DFS(x))
break;
int d=inf;
for(int y=0;y<ny;y++)
if(!visy[y]&&d>slack[y])
d=slack[y];
for(int i=0;i<nx;i++)
if(visx[i])
lx[i]-=d;
for(int i=0;i<ny;i++)
{
if(visy[i])
ly[i]+=d;
else
slack[i]-=d;
}
}
}
int ans=0;
for(int y=0;y<ny;y++)
{
if(linked[y]!=-1)
ans+=g[linked[y]][y];
}
return ans;
} int main()
{
int n;
while(scanf("%d",&n)!=EOF&&n)
{
nx=ny=n;
for(int i=0;i<n;i++)
scanf("%d",&v[i]);
int ch;
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
scanf("%1d",&ch);//输入格式1d
if(ch==1)
g[i][j]=v[i]^v[j];
else
g[i][j]=0;
}
}
printf("%d\n",KM());
}
return 0;
}

版权声明:本文博主原创文章,博客,未经同意不得转载。

[ACM] HDU 3395 Special Fish (最大重量二分图匹配,KM算法)的更多相关文章

  1. 训练指南 UVALive - 4043(二分图匹配 + KM算法)

    layout: post title: 训练指南 UVALive - 4043(二分图匹配 + KM算法) author: "luowentaoaa" catalog: true ...

  2. HDU 3395 Special Fish(拆点+最大费用最大流)

    Special Fish Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Tot ...

  3. 牛客多校第五场 E room 二分图匹配 KM算法模板

    链接:https://www.nowcoder.com/acm/contest/143/E来源:牛客网 Nowcoder University has 4n students and n dormit ...

  4. 二分图匹配--KM算法

    Kuhn-Munkres算法 KM算法,求完备匹配下的最大权匹配,时间复杂度O(\(n^3\)) 所谓的完备匹配就是在二部图中,x点集中的所有点都有对应的匹配 且 y点集中所有的点都有对应的匹配 ,则 ...

  5. ACM学习历程—POJ3565 Ants(最佳匹配KM算法)

    Young naturalist Bill studies ants in school. His ants feed on plant-louses that live on apple trees ...

  6. HDU 3395 Special Fish 最“大”费用最大流

    求最大费用能够将边权取负以转化成求最小费用. 然而此时依旧不正确.由于会优先寻找最大流.可是答案并不一定出如今满流的时候.所以要加一些边(下图中的红边)使其在答案出现时满流. 设全部边的流量为1,花费 ...

  7. HDU 5943 Kingdom of Obsession 【二分图匹配 匈牙利算法】 (2016年中国大学生程序设计竞赛(杭州))

    Kingdom of Obsession Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Oth ...

  8. 【HDU 2255】奔小康赚大钱 (最佳二分匹配KM算法)

    奔小康赚大钱 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Subm ...

  9. USACO 4.2 The Perfect Stall(二分图匹配匈牙利算法)

    The Perfect StallHal Burch Farmer John completed his new barn just last week, complete with all the ...

随机推荐

  1. Selenium来抓取动态加载的页面

    一般的爬虫都是直接使用http协议,下载指定url的html内容,并对内容进行分析和抽取.在我写的爬虫框架webmagic里也使用了HttpClient来完成这样的任务. 但是有些页面是通过js以及a ...

  2. 最短路径算法-Dijkstra算法的应用之单词转换(词梯问题)(转)

    一,问题描述 在英文单词表中,有一些单词非常相似,它们可以通过只变换一个字符而得到另一个单词.比如:hive-->five:wine-->line:line-->nine:nine- ...

  3. [置顶] github简单使用

    git的介绍可以看这里   http://zh.wikipedia.org/wiki/GitHub 安装和使用参考的这个    http://www.cnblogs.com/cocowool/arch ...

  4. hdu4513(manacher)

    传送门:吉哥系列故事——完美队形II 题意:求最长回文队伍且队伍由中间向两边递减. 分析:manach算法小应用,在判断回文子串向两边递减时加点限制使回文是由中间向两边递减的. #pragma com ...

  5. 【 D3.js 入门系列 --- 8 】 对话操作(事件)

    本人的个人博客为: www.ourd3js.com csdn博客为: blog.csdn.net/lzhlzz 转载请注明出处,谢谢. 这一节介绍怎样进行对话的操作,如鼠标单击,鼠标滑过等. 对一个被 ...

  6. POJ 2151 Check the difficulty of problems (动态规划-可能DP)

    Check the difficulty of problems Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4522   ...

  7. 使用WiX Toolset创建.NET程序发布Bootstrapper(安装策略管理)(二)——自定义安装

    原文:使用WiX Toolset创建.NET程序发布Bootstrapper(安装策略管理)(二)--自定义安装 自定义产品卸载方式 继续从上一次的基础上前进,现在我们已经知道了最简单的bootstr ...

  8. CCLuaObjcBridge调Objective-C方法传索引数组报invalid key to &#39;next&#39;错调试

    CCLuaObjcBridge是cocos2d-x系列引擎与Objective-C进行交互的"桥梁",老廖的quick-cocos2d-x在其framework进行了简单了封装,封 ...

  9. SVN的svnlook命令

    svnlook命令集(zhuanzai) 2011-12-08 17:00:30|  分类: System and CVS|字号 订阅     svnlook 名称 svnlook author — ...

  10. hdu1166(线段树)

    题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1166 线段树功能:update:单点增减 query:区间求和 #pragma comment(lin ...