POJ 2711 Leapin' Lizards / HDU 2732 Leapin' Lizards / BZOJ 1066 [SCOI2007]蜥蜴(网络流,最大流)

Description

Your platoon of wandering lizards has entered a strange room in the labyrinth you are exploring. As you are looking around for hidden treasures, one of the rookies steps on an innocent-looking stone and the room's floor suddenly disappears! Each lizard in your platoon is left standing on a fragile-looking pillar, and a fire begins to rage below...

Leave no lizard behind! Get as many lizards as possible out of the room, and report the number of casualties.

The pillars in the room are aligned as a grid, with each pillar one unit away from the pillars to its east, west, north and south. Pillars at the edge of the grid are one unit away from the edge of the room (safety). Not all pillars necessarily have a lizard. A lizard is able to leap onto any unoccupied pillar that is within d units of his current one. A lizard standing on a pillar within leaping distance of the edge of the room may always leap to safety... but there's a catch: each pillar becomes weakened after each jump, and will soon collapse and no longer be usable by other lizards. Leaping onto a pillar does not cause it to weaken or collapse; only leaping off of it causes it to weaken and eventually collapse. Only one lizard may be on a pillar at any given time.

Input

The input file will begin with a line containing a single integer representing the number of test cases, which is at most 25. Each test case will begin with a line containing a single positive integer n representing the number of rows in the map, followed by a single non-negative integer d representing the maximum leaping distance for the lizards. Two maps will follow, each as a map of characters with one row per line. The first map will contain a digit (0-3) in each position representing the number of jumps the pillar in that position will sustain before collapsing (0 means there is no pillar there). The second map will follow, with an 'L' for every position where a lizard is on the pillar and a '.' for every empty pillar. There will never be a lizard on a position where there is no pillar.

Each input map is guaranteed to be a rectangle of size n x m, where 1 <= n <= 20 and 1 <= m <= 20. The leaping distance is always 1 <= d <= 3.

Output

For each input case, print a single line containing the number of lizards that could not escape. The format should follow the samples provided below.

Sample Input

4

3 1

1111

1111

1111

LLLL

LLLL

LLLL

3 2

00000

01110

00000

.....

.LLL.

.....

3 1

00000

01110

00000

.....

.LLL.

.....

5 2

00000000

02000000

00321100

02000000

00000000

........

........

..LLLL..

........

........

Sample Output

Case #1: 2 lizards were left behind.

Case #2: no lizard was left behind.

Case #3: 3 lizards were left behind.

Case #4: 1 lizard was left behind.

Http

POJ:https://vjudge.net/problem/POJ-2711

HDU:https://vjudge.net/problem/HDU-2732

BZOJ:http://www.lydsy.com/JudgeOnline/problem.php?id=1066

Source

网络流,最大流

题目大意

在一个迷宫中有若干只蜥蜴,每只蜥蜴一次可以跳跃给定距离(曼哈顿距离),每一个格子都有一个能承载的上限,最多只能有这么多只蜥蜴跳到这个上面(即以之为中转点)。一个格子在同一时间只能有一只蜥蜴。若蜥蜴能调到迷宫外则称之为逃离出去了。现在求最少几只蜥蜴不能跳出去(即使得最多的蜥蜴逃出迷宫)

解决思路

首先关于每个点的处理,因为每一格能通过的蜥蜴的数量是有限的,所以我们可以拆点,来限制通过每一点的容量。

然后,如果有一个点初始有蜥蜴,则从源点出发连一条边。若某个点可以直接跳到迷宫外面,则连一条边到汇点。

最后对于从一个点可以跳到的点,连容量为无穷大的边。

然后跑一边最大流即可求出获救的蜥蜴数量。

注意输出格式,如no和是否有复数形式。

这里使用Dinic实现最大流,可以参考这篇文章

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std; //这两个分别是把二维坐标转成一维数组地址,和把一维数组地址转成坐标输出
#define pos(x,y) ((x-1)*m+y)
#define inpos(x) '['<<x<<']'<<'('<<(x/m)+(int)(!(x%m==0))<<','<<(x-1)%m+1<<')' const int maxN=2001;
const int maxM=maxN*maxN*2;
const int inf=2147483647; class Edge
{
public:
int u,v,flow;
}; int n,m;//即题目中的n*m矩阵
int D;//蜥蜴能跳的最长距离
int Lizards;//蜥蜴数
int cnt;
int Head[maxN];
int Next[maxM];
Edge E[maxM];
int depth[maxN];
int cur[maxN];
int Q[maxN];
char str1[maxN][maxN];//存每个格子能跳过的最多蜥蜴数
char str[maxN]; void Add_Edge(int u,int v,int flow);
bool bfs();
int dfs(int u,int flow); int main()
{
int T;
scanf("%d",&T);
for (int ti=1;ti<=T;ti++)
{
Lizards=0;//多组数据,先初始化
cnt=-1;
memset(Head,-1,sizeof(Head));
scanf("%d%d",&n,&D);
cin>>str1[1];//先读入一个,得出m,方便处理
m=strlen(str1[1]);
for (int i=1;i<=n;i++)
{
for (int j=0;j<m;j++)
if (str1[i][j]!='0')
Add_Edge(pos(i,j+1),pos(i,j+1)+n*m,str1[i][j]-'0');//连接拆点后的两个点
if (i!=n)
cin>>str1[i+1];
}
for (int i=1;i<=n;i++)
{
cin>>str;
for (int j=0;j<m;j++)
if (str[j]=='L')
{
Lizards++;
Add_Edge(0,pos(i,j+1),1);//连接蜥蜴与源点
}
}
/*
for (int i=1;i<=n*m*2;i++)
{
for (int j=Head[i];j!=-1;j=Next[j])
if (E[j].v!=flow)
cout<<inpos(i)<<"->"<<inpos(E[j].v)<<" "<<E[j].flow<<endl;
}
//*/
for (int i=1;i<=n;i++)//连接每一个格子
for (int j=1;j<=m;j++)
if (str1[i][j-1]!='0')
{
if ((i<=D)||(j<=D)||(n-i+1<=D)||(m-j+1<=D))
{
Add_Edge(pos(i,j)+n*m,n*m*2+1,inf);//连接能跳出迷宫的格子与汇点
//continue;
}
for (int x=max(i-D,1);x<=min(i+D,n);x++)//连接能够跳到的格子
for (int y=max(j-D,1);y<=min(j+D,m);y++)
{
if ((x==i)&&(j==y))
continue;
if (str1[x][y-1]=='0')
continue;
if (abs(x-i)+abs(y-j)<=D)//主语是曼哈顿距离
Add_Edge(pos(i,j)+n*m,pos(x,y),inf);
}
}
/*
for (int i=0;i<=n*m*2;i++)
{
for (int j=Head[i];j!=-1;j=Next[j])
if (E[j].flow!=0)
cout<<inpos(i)<<"->"<<inpos(E[j].v)<<" "<<E[j].flow<<endl;
}
//*/
int Ans=0;//求解最大流
while (bfs())
{
for (int i=0;i<=n*m*2+1;i++)
cur[i]=Head[i];
while (int di=dfs(0,inf))
Ans+=di;
}
Ans=Lizards-Ans;//因为要求的是不能跳出去的蜥蜴的个数,所以要用总数减去最大流
printf("Case #%d: ",ti);
if (Ans==0)//注意输出格式,如果是0输出no,如果是1输出单数形式,如果大于1输出复数形式
printf("no lizard was left behind.\n");
else
if (Ans==1)
printf("1 lizard was left behind.\n");
else
printf("%d lizards were left behind.\n",Ans);
//printf("%d\n",Ans);
}
return 0;
} void Add_Edge(int u,int v,int flow)
{
cnt++;
Next[cnt]=Head[u];
Head[u]=cnt;
E[cnt].u=u;
E[cnt].v=v;
E[cnt].flow=flow; cnt++;
Next[cnt]=Head[v];
Head[v]=cnt;
E[cnt].u=v;
E[cnt].v=u;
E[cnt].flow=0;
} bool bfs()
{
memset(depth,-1,sizeof(depth));
int h=1,t=0;
Q[1]=0;
depth[0]=1;
do
{
t++;
int u=Q[t];
for (int i=Head[u];i!=-1;i=Next[i])
{
int v=E[i].v;
if ((depth[v]==-1)&&(E[i].flow>0))
{
depth[v]=depth[u]+1;
h++;
Q[h]=v;
}
}
}
while (t!=h);
if (depth[n*m*2+1]==-1)
return 0;
return 1;
} int dfs(int u,int flow)
{
if (u==n*m*2+1)
return flow;
for (int i=Head[u];i!=-1;i=Next[i])
{
int v=E[i].v;
if ((depth[v]==depth[u]+1)&&(E[i].flow>0))
{
int di=dfs(v,min(flow,E[i].flow));
if (di>0)
{
E[i].flow-=di;
E[i^1].flow+=di;
return di;
}
}
}
return 0;
}

POJ 2711 Leapin' Lizards / HDU 2732 Leapin' Lizards / BZOJ 1066 [SCOI2007]蜥蜴(网络流,最大流)的更多相关文章

  1. poj 2711 Leapin' Lizards && BZOJ 1066: [SCOI2007]蜥蜴 最大流

    题目链接:http://poj.org/problem?id=2711 题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1066 Your p ...

  2. 【解题报告】 Leapin' Lizards HDU 2732 网络流

    [解题报告] Leapin' Lizards HDU 2732 网络流 题外话 在正式讲这个题目之前我想先说几件事 1. 如果大家要做网络流的题目,我在网上看到一个家伙,他那里列出了一堆网络流的题目, ...

  3. HDU 2732 Leapin&#39; Lizards(拆点+最大流)

    HDU 2732 Leapin' Lizards 题目链接 题意:有一些蜥蜴在一个迷宫里面,有一个跳跃力表示能跳到多远的柱子,然后每根柱子最多被跳一定次数,求这些蜥蜴还有多少是不管怎样都逃不出来的. ...

  4. hdu 2732 Leapin' Lizards(最大流)Mid-Central USA 2005

    废话: 这道题不难,稍微构造一下图就可以套最大流的模板了.但是我还是花了好久才解决.一方面是最近确实非常没状态(托词,其实就是最近特别颓废,整天玩游戏看小说,没法静下心来学习),另一方面是不够细心,输 ...

  5. HDU 2732 Leapin' Lizards(最大流)

    http://acm.hdu.edu.cn/showproblem.php?pid=2732 题意: 给出n行的网格,还有若干只蜥蜴,每只蜥蜴一开始就在一个格子之中,并且给出蜥蜴每次的最大跳跃长度d. ...

  6. Leapin' Lizards(hdu 2732)

    Leapin' Lizards Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)T ...

  7. K - Leapin' Lizards - HDU 2732(最大流)

    题意:在一个迷宫里面有一些蜥蜴,这个迷宫有一些柱子组成的,并且这些柱子都有一个耐久值,每当一只蜥蜴跳过耐久值就会减一,当耐久值为0的时候这个柱子就不能使用了,每个蜥蜴都有一个最大跳跃值d,现在想知道有 ...

  8. HDU 2732 Leapin' Lizards

    网络最大流+拆点.输出有坑!!! #include<cstdio> #include<cstring> #include<string> #include<c ...

  9. Leapin' Lizards HDU - 2732 (恶心的建图。。)

    这道题其实不难...就是建图恶心了点....emm... 题意: 多源多汇 + 拆边 青蛙跳柱子, 每根柱子都有一定的承载能力, 青蛙跳上去之后柱子的承载能力就会减一,跳到边界就能活 跳不到就over ...

随机推荐

  1. 聊聊Zookeeper应用场景、架构设计、选主机制

    Zookeeper作为一个分布式协调系统提供了一项基本服务:分布式锁服务,分布式锁是分布式协调技术实现的核心内容.像配置管理.任务分发.组服务.分布式消息队列.分布式通知/协调等,这些应用实际上都是基 ...

  2. 个人java框架 技术分析

    1.框架选型 spring-boot https://github.com/JeffLi1993/springboot-learning-example https://mp.weixin.qq.co ...

  3. mysql操作命令梳理(4)-grant授权和revoke回收权限

    在mysql维护工作中,做好权限管理是一个很重要的环节.下面对mysql权限操作进行梳理: mysql的权限命令是grant,权限撤销的命令时revoke:grant授权格式:grant 权限列表 o ...

  4. Spring RPC 入门学习(3)-获取Student对象

    Spring RPC传递对象. 1. 新建RPC接口:StudentInterface.java package com.cvicse.ump.rpc.interfaceDefine; import ...

  5. python基础学习笔记(五)

    字符串基本操作 所有标准的序列操作(索引.分片.乘法.判断成员资格.求长度.取最小值和最大值)对字符串同样适用,前面已经讲述的这些操作.但是,请注意字符串都是不可变的. 字符串的方法: 字符串从str ...

  6. HDU-6440-费马小定理

    亏我前几天还学数论呢...没有深入研究费马小定理这个东西...做事情一定要静下心来啊... 题目要求满足(m+n)^p=m^p+n^p,要你定义一个封闭的新的加法和乘法运算 我们知道费马小定理中有两种 ...

  7. 研究C语言的新型编译环境TCC

    C语言综合研究1 搭建一个tcc环境 研究过程: 问题引出:为什么要使用tcc环境,甚至连图形界面都没有,为什么要使用这样的化境? 按照我们学习的本质来讲,可能是为了体验C语言底层的相关特性,但是在研 ...

  8. 2-Twenty First Scrum Meeting-20151221

    任务安排 成员 今日完成 明日任务 闫昊 请假(数据库)   唐彬 请假(数据库)   史烨轩  尝试使用downloadmanager对notification进行更新  尝试使用downloadm ...

  9. git常用命令点击查看

    创建git项目仓库 $git init 配置个人登记信息,这样团队协作的时候,就可以看到哪个用户修改过哪些文件的 $git config --global user.name 'cfanbo' $gi ...

  10. Golang的格式化输出fmt.Printf

    本文来源:Go by example. Golang的格式化输出 和 C语言的标准输出基本一样,但是增加了一些针对Golang语言的特有数据结构的格式化输出方式. 一下就是实例: package ma ...