HDU-2732-leapin'Lizards(最大流, 拆点)
链接:
https://vjudge.net/problem/HDU-2732
题意:
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.
思路:
依然是拆点建图.每个柱子拆成入口和出口,权值为可用次数,出口连到别的点和边缘都为INF.
对每个有人的柱子,源点连一个权值为1的边.
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 20+10;
const int INF = 1e9;
struct Edge
{
int from, to, cap;
};
vector<Edge> edges;
vector<int> G[MAXN*MAXN*4];
int Dis[MAXN*MAXN*4];
char Map1[MAXN][MAXN], Map2[MAXN][MAXN];
int n, m, s, t, d;
void AddEdge(int from, int to, int cap)
{
edges.push_back(Edge{from, to, cap});
edges.push_back(Edge{to, from, 0});
G[from].push_back(edges.size()-2);
G[to].push_back(edges.size()-1);
}
bool Bfs()
{
memset(Dis, -1, sizeof(Dis));
queue<int> que;
que.push(s);
Dis[s] = 0;
while (!que.empty())
{
int u = que.front();
que.pop();
// cout << u << endl;
for (int i = 0;i < G[u].size();i++)
{
Edge &e = edges[G[u][i]];
if (e.cap > 0 && Dis[e.to] == -1)
{
Dis[e.to] = Dis[u]+1;
que.push(e.to);
}
}
}
return Dis[t] != -1;
}
int Dfs(int u, int flow)
{
if (u == t)
return flow;
int res = 0;
for (int i = 0;i < G[u].size();i++)
{
Edge &e = edges[G[u][i]];
if (e.cap > 0 && Dis[u]+1 == Dis[e.to])
{
int tmp = Dfs(e.to, min(flow, e.cap));
// cout << "flow:" << e.from << ' ' << e.to << ' ' << tmp << endl;
e.cap -= tmp;
flow -= tmp;
edges[G[u][i]^1].cap += tmp;
res += tmp;
if (flow == 0)
break;
}
}
if (res == 0)
Dis[u] = -1;
return res;
}
int MaxFlow()
{
int res = 0;
while (Bfs())
{
res += Dfs(s, INF);
}
return res;
}
int main()
{
// freopen("test.in", "r", stdin);
ios::sync_with_stdio(false);
cin.tie(0);
int T, cnt = 0;
cin >> T;
while (T--)
{
cin >> n >> d;
for (int i = 1;i <= n;i++)
cin >> (Map1[i]+1);
for (int i = 1;i <= n;i++)
cin >> (Map2[i]+1);
m = strlen(Map1[1]+1);
s = 0, t = n*m*2+1;
for (int i = s;i <= t;i++)
G[i].clear();
edges.clear();
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= m;j++)
{
//(i-1)*m+j
if (Map1[i][j] == 0)
continue;
int node = (i-1)*m+j;
AddEdge(node*2-1, node*2, Map1[i][j]-'0');
if (i <= d || i > n-d || j <= d || j > m-d)
{
// cout << i << ' ' << j << endl;
AddEdge(node*2, t, INF);
}
for (int z = 1;z <= n;z++)
{
for (int k = 1;k <= m;k++)
{
if (abs(z-i)+abs(k-j) > d)
continue;
if (i == z && j == k)
continue;
// cout << i << ' ' << j << ' ' << ' ' << z << ' ' << k << endl;
int nodeto = (z-1)*m+k;
AddEdge(node*2, nodeto*2-1, INF);
// cout << Map1[i][j]-'0' << endl;
}
}
}
}
int res = 0;
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= m;j++)
{
if (Map2[i][j] == '.')
continue;
res++;
int node = (i-1)*m+j;
AddEdge(s, node*2-1, 1);
}
}
res -= MaxFlow();
if (res == 0)
cout << "Case #" << ++cnt << ": no lizard was left behind." << endl;
else if (res == 1)
cout << "Case #" << ++cnt << ": " << res << " lizard was left behind." << endl;
else
cout << "Case #" << ++cnt << ": " << res << " lizards were left behind." << endl;
}
return 0;
}
/*
10
3 1
111
111
111
LLL
LLL
LLL
*/
HDU-2732-leapin'Lizards(最大流, 拆点)的更多相关文章
- hdu 2732 Leapin' Lizards 最大流 拆点 建图
题目链接 题意 给定一张网格,格子中有些地方有柱子,有些柱子上面有蜥蜴. 每个柱子只能承受有限只蜥蜴从上面经过.每只蜥蜴每次能走到相距曼哈顿距离\(\leq k\)的格子中去. 问有多少只蜥蜴能走出网 ...
- hdu 2732 Leapin' Lizards (最大流 拆点建图)
Problem Description Your platoon of wandering lizards has entered a strange room in the labyrinth yo ...
- POJ 2711 Leapin' Lizards / HDU 2732 Leapin' Lizards / BZOJ 1066 [SCOI2007]蜥蜴(网络流,最大流)
POJ 2711 Leapin' Lizards / HDU 2732 Leapin' Lizards / BZOJ 1066 [SCOI2007]蜥蜴(网络流,最大流) Description Yo ...
- HDU 2732 Leapin' Lizards
网络最大流+拆点.输出有坑!!! #include<cstdio> #include<cstring> #include<string> #include<c ...
- HDU - 2732 Leapin' Lizards (拆点最大流)
题意:有N*M的矩形,每个格点有一个柱子,每根柱子有高度c,允许蜥蜴经过这根柱子c次,开始有一些蜥蜴在某些柱子上,它们要跳出这个矩形,每步最大能跳d个单位,求最少有多少蜥蜴不能跳出这个矩形. 分析:转 ...
- 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. ...
- 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 ...
- hdu2732 Leapin' Lizards 最大流+拆点
Your platoon of wandering lizards has entered a strange room in the labyrinth you are exploring. As ...
- HDU 2732 Leapin' 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. ...
- HDU2732 Leapin' Lizards —— 最大流、拆点
题目链接:https://vjudge.net/problem/HDU-2732 Leapin' Lizards Time Limit: 2000/1000 MS (Java/Others) M ...
随机推荐
- java:struts框架2(方法的动态和静态调用,获取Servlet API三种方式(推荐IOC(控制反转)),拦截器,静态代理和动态代理(Spring AOP))
1.方法的静态和动态调用: struts.xml: <?xml version="1.0" encoding="UTF-8"?> <!DOCT ...
- java:异常机制(try,catch,finally,throw,throws,自定义异常)
* String类中的格式化字符串的方法: * public static String format(String format, Object... args):使用指定的格式字符串和参数返回一个 ...
- Java编程思想—八皇后问题(数组法、堆栈法)
Java编程思想-八皇后问题(数组法.堆栈法) 实验题目:回溯法实验(八皇后问题) 实验目的: 实验要求: 实验内容: (1)问题描述 (2)实验步骤: 数组法: 堆栈法: 算法伪代码: 实验结果: ...
- 【转载】GitHub 标星 1.2w+,超全 Python 常用代码合集,值得收藏!
本文转自逆袭的二胖,作者二胖 今天给大家介绍一个由一个国外小哥用好几年时间维护的 Python 代码合集.简单来说就是,这个程序员小哥在几年前开始保存自己写过的 Python 代码,同时把一些自己比较 ...
- CSS基本样式-文本属性
字体属性 文本属性呢,自我认为就是写文档的一些格式属性,例如:字体颜色,字体加粗,字体大小,字体类型等,而且我们在输出测试用例报告的时候也可以用到这些属性,对测试报告进行优化. <html> ...
- HDU 1042 N!(高精度阶乘、大数乘法)
N! Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submi ...
- [Python3] 025 包
目录 1. 模块 1.1 模块是什么? 1.2 为什么用模块? 1.3 如何定义模块? 1.4 如何使用模块? 1.4.1 例子1 1.4.2 例子2 1.4.3 例子3 1.4.4 例子4 1.4. ...
- 搜索专题:Balloons
搜索专题:Balloons 这道题一看与时间有关,第一想到的就是BFS,定义一个状态,包含每一个状态的剩余气球数,已经进行的时间和每一个志愿者上一次吹气球的时间: 每一次状态转换时,检查是否有没有使用 ...
- 【二】Django 视图和url配置
在新建的Django项目下,新建一个views的python文件,编辑如下代码 from django.http import HttpResponse def hello(request): ret ...
- 关于jQuery获取不到动态添加的元素节点的问题
遇到问题: 当我获取 $("#art-list")页面元素后去在后面追加标签的时候(append),在下面用 $(selector) 获取刚刚添加的标签,发现怎么都获取不到. 问题 ...