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HDU 2732 Leapin' Lizards(拆点+最大流)
题目意思是有一些蜥蜴在一个迷宫里面，求这些蜥蜴还有多少是无论如何都逃不出来的。题目只给定一个行数n，一个最远能够跳跃的距离d。每只蜥蜴有一个初始的位置，题目保证这些位置都有一些柱子，但是它每离开一根柱子，柱子的高度就会降低1m，问最多能有多少只跳不出去。
将每个柱子在的点进行拆点，把每一个点拆完之后连一条容量为所在点柱子高度的边。从原点连一条容量为1的边，然后找到每个可以直接跳出的点，将这些点与汇点 相连容量为无穷。每个柱子与它可以到达的点的容量也为无穷。
Leapin‘ Lizards
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1254 Accepted Submission(s): 524
Problem 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.
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.
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.
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-12
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
using namespace std;
const int maxn = 1100;
int cnt;
int n, m;
int cur[maxn], head[maxn];
int dis[maxn], gap[maxn];
int aug[maxn], pre[maxn];
int num[maxn];
struct node
{
int v, w;
int next;
} f[2010000];
void init()
{
cnt = 0;
memset(head, -1, sizeof(head));
}
void add(int u, int v, int w)
{
f[cnt].v = v;
f[cnt].w = w;
f[cnt].next = head[u];
head[u] = cnt++;
f[cnt].v = u;
f[cnt].w = 0;
f[cnt].next = head[v];
head[v] = cnt++;
}
int SAP(int s, int e, int n)
{
int max_flow = 0, v, u = s;
int id, mindis;
aug[s] = INF;
pre[s] = -1;
memset(dis, 0, sizeof(dis));
memset(gap, 0, sizeof(gap));
gap[0] = n;
for (int i = 0; i <= n; ++i) cur[i] = head[i];/// 初始化当前弧为第一条弧
while (dis[s] < n)
{
bool flag = false;
if (u == e)
{
max_flow += aug[e];
for (v = pre[e]; v != -1; v = pre[v]) /// 路径回溯更新残留网络
{
id = cur[v];
f[id].w -= aug[e];
f[id^1].w += aug[e];
aug[v] -= aug[e]; /// 修改可增广量，以后会用到
if (f[id].w == 0) u = v; /// 不回退到源点，仅回退到容量为0的弧的弧尾
}
}
for (id = cur[u]; id != -1; id = f[id].next)/// 从当前弧开始查找允许弧
{
v = f[id].v;
if (f[id].w > 0 && dis[u] == dis[v] + 1) /// 找到允许弧
{
flag = true;
pre[v] = u;
cur[u] = id;
aug[v] = min(aug[u], f[id].w);
u = v;
break;
}
}
if (flag == false)
{
if (--gap[dis[u]] == 0) break; ///gap优化，层次树出现断层则结束算法
mindis = n;
cur[u] = head[u];
for (id = head[u]; id != -1; id = f[id].next)
{
v = f[id].v;
if (f[id].w > 0 && dis[v] < mindis)
{
mindis = dis[v];
cur[u] = id; /// 修改标号的同时修改当前弧
}
}
dis[u] = mindis + 1;
gap[dis[u]]++;
if (u != s) u = pre[u]; /// 回溯继续寻找允许弧
}
}
return max_flow;
}
char map1[maxn][maxn], map2[maxn][maxn];
int vis[maxn][maxn];
double dist(int x1, int y1, int x2, int y2)
{
double a = x1, b = y1, c = x2, d = y2;
return sqrt((a-c)*(a-c)+(b-d)*(b-d));
}
int main()
{
int Case = 1;
int d;
int K;
cin >>K;
while(K--)
{
scanf("%d %d",&n, &d);
init();
memset(vis, 0, sizeof(vis));
for(int i = 0; i < n; i++) cin >>map1[i];
for(int j = 0; j < n; j++) cin >>map2[j];
int len = strlen(map1[0]);
int k = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < len; j++) if(map1[i][j]-'0' > 0) vis[i][j] = ++k;
int S = 0;
int T = 2*k+1;
int en = T+1;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < len; j++)
{
if(map1[i][j]-'0' > 0)
{
add(vis[i][j], vis[i][j]+k, map1[i][j]-'0');
for(int ii = 0; ii < n; ii++)
{
for(int jj = 0; jj < len; jj++)
{
if(i == ii && j == jj) continue;
double s = dist(i, j, ii, jj);
if(vis[ii][jj] && (double)d >= s) add(vis[i][j]+k, vis[ii][jj], INF-10);
}
}
}
}
}
int kk = 0;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < len; j++)
{
if(map2[i][j] == 'L')
{
kk++;
add(S, vis[i][j], 1);
}
}
}
for(int i = 0; i < n; i++)
for(int j = 0; j < len; j++)
if(map1[i][j]-'0' > 0) if(i+1<=d || j+1<=d || n-i<=d || len-j<=d) add(vis[i][j]+k, T, INF-10);
int ans = SAP(S, T, en);
cout<<"Case #"<<Case++<<": ";
if(kk-ans == 0) cout<<"no lizard was left behind."<<endl;
else if(kk-ans == 1) cout<<"1 lizard was left behind."<<endl;
else cout<<kk-ans<<" lizards were left behind."<<endl;
}
return 0;
}
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