题意：
每一个' . '有一个姑娘， E是出口，'.'是空地 , 'X‘ 是墙。
每秒钟每一个姑娘能够走一步（上下左右）
每秒钟每一个出口仅仅能出去一个人
给定n*m的地图， 时限T
问全部姑娘是否能在T秒内逃生，若能输出最小值，不能输出"impossible"
思路：
显然是二分答案+网络流判可行。
由于每一个出口每秒钟仅仅能出去一个人，那么就把每一个出口按时间拆点，则T秒钟就拆成T个点。
网络流建图
1、源点 到 每一个姑娘 建流量为1的边。
2、若某姑娘到 a出口须要时间为 t秒，则建一条流量为1的边 连向a出口拆点为t秒的点。
3、每一个出口的全部拆点向汇点连一条流量为1的边。
4、对于每一个出口u的x秒拆点，向u的x+1秒的拆点连一条流量为inf的边（表示从x秒来的人假设x秒还从u出不去，能够在u等到x+1秒出去）
二分一下 第二点中的 t 秒(即答案），推断最大流是否等于人数
二分匹配建图：
枚举时间（由于时间最大仅仅有12*12）
在原图的基础上加上以下的边：
在i时间内若能出去则那姑娘向全部能出去的 出口时间拆点连边。
再在原来的匹配上继续增广，累计最大匹配数。
当最大匹配数==人数时则是最小的时间。
数据较水能够用点非主流的建图卡过去，预计仅仅有30组数据。
网络流代码：
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define ll int
#define N 20050
#define M 105000
#define inf 10737418
struct Edge{
ll from, to, cap, nex;
}edge[M*4];//注意这个一定要够大 不然会re 还有反向弧
ll head[N], edgenum;
void add(ll u, ll v, ll cap){
Edge E = { u, v, cap, head[u]};
edge[ edgenum ] = E;
head[u] = edgenum ++;
Edge E2= { v, u, 0, head[v]};
edge[ edgenum ] = E2;
head[v] = edgenum ++;
}
ll sign[N];
bool BFS(ll from, ll to){
memset(sign, -1, sizeof(sign));
sign[from] = 0;
queue<ll>q;
q.push(from);
while( !q.empty() ){
int u = q.front(); q.pop();
for(ll i = head[u]; i!=-1; i = edge[i].nex)
{
ll v = edge[i].to;
if(sign[v]==-1 && edge[i].cap)
{
sign[v] = sign[u] + 1, q.push(v);
if(sign[to] != -1)return true;
}
}
}
return false;
}
ll Stack[N], top, cur[N];
ll dinic(ll from, ll to){
ll ans = 0;
while( BFS(from, to) )
{
memcpy(cur, head, sizeof(head));
ll u = from; top = 0;
while(1)
{
if(u == to)
{
ll flow = inf, loc;//loc 表示 Stack 中 cap 最小的边
for(ll i = 0; i < top; i++)
if(flow > edge[ Stack[i] ].cap)
{
flow = edge[Stack[i]].cap;
loc = i;
}
for(ll i = 0; i < top; i++)
{
edge[ Stack[i] ].cap -= flow;
edge[Stack[i]^1].cap += flow;
}
ans += flow;
top = loc;
u = edge[Stack[top]].from;
}
for(ll i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标
if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;
if(cur[u] != -1)
{
Stack[top++] = cur[u];
u = edge[ cur[u] ].to;
}
else
{
if( top == 0 )break;
sign[u] = -1;
u = edge[ Stack[--top] ].from;
}
}
}
return ans;
}
void init(){memset(head,-1,sizeof head);edgenum = 0;}
char mp[15][15];
int n, m, T;
int Hash(int x,int y){return x*m+y;}
vector<int>E,P;
int dis[150][150], step[4][2]={1,0,-1,0,0,1,0,-1};
bool vis[150][150];
void bfs(int sx,int sy){
int start = Hash(sx,sy);
memset(vis, 0, sizeof vis);
vis[sx][sy] = 1;
dis[start][start] = 0;
queue<int>qx,qy; while(!qx.empty())qx.pop(); while(!qy.empty())qy.pop();
qx.push(sx), qy.push(sy);
while(!qx.empty()){
int x = qx.front(), y = qy.front();
qx.pop(); qy.pop();
for(int i = 0; i < 4; i++){
int dx = x + step[i][0], dy = y + step[i][1];
if(!(0<=dx&&dx<n&&0<=dy&&dy<m))continue;
if(vis[dx][dy] || mp[dx][dy]!='.')continue;
vis[dx][dy] = 1;
dis[Hash(dx,dy)][start] = dis[start][Hash(dx,dy)] = dis[start][Hash(x,y)]+1;
qx.push(dx); qy.push(dy);
}
}
}
bool ok(int TIME){
init();
int from = N-2, to = N-1;
for(int i = 0; i < P.size(); i++)add(from, P[i], 1);
for(int i = 0; i < P.size(); i++)
{
for(int j = 0; j < E.size(); j++)
if(dis[P[i]][E[j]]<=TIME) add(P[i],j*150+150+dis[P[i]][E[j]],1);
}
for(int i = 0; i < E.size(); i++)
for(int j = 1; j <= TIME; j++)
{
add(i*150+150+j, to, 1);
if(j!=TIME)
add(i*150+150+j,i*150+150+j+1,inf);
}
return dinic(from,to)==P.size();
}
int main(){
//freopen("date.in","r+",stdin);
//freopen("ans.out","w+",stdout);
int i, j;
while(~scanf("%d %d %d",&n,&m,&T)){
E.clear(); P.clear();
memset(dis, 0, sizeof dis);
memset(mp, 0, sizeof mp);
for(i=0;i<n;i++)scanf("%s",mp[i]);
for(i=0;i<n;i++)for(j=0;j<m;j++)
{
if(mp[i][j]=='E')E.push_back(Hash(i,j));
else if(mp[i][j]=='.')P.push_back(Hash(i,j));
}
if(P.size()==0){puts("0");continue;} if(E.size()==0){puts("impossible");continue;}
for(i = 0; i < E.size(); i++)bfs(E[i]/m, E[i]%m);
int l = 1, r = 256, ans = inf;
while(l<=r)
{
int mid = (l+r)>>1;
if(ok(mid))ans = min(ans, mid), r = mid-1;
else l = mid+1;
}
if(T<ans || ans==inf){puts("impossible");continue;}
else printf("%d
",ans);
}
return 0;
}
/*
6 12 100000
......E....E
E.EEE...E...
...E....EE..
.E.E.......E
.....E......
.E...E...EE.
11 3 100000
X..
...
.E.
EE.
E.E
...
...
.E.
E..
..E
E..
7 6 100000
.E..E.
...E..
......
...E.E
...X..
...EE.
....E.
11 10 100000
.E........
......E...
........E.
EE....E..E
...E...E..
......XE.E
..........
.........X
.........E
.EE.......
..EE.E.E.E
7 3 100000
...
..E
...
..X
EE.
...
.X.
3 2 100000
..
..
EE
6 7 100000
..E....
.......
.....E.
...EE..
.......
.......
6 8 100000
..EEEE..
...E....
........
........
.......E
E......E
10 8 100000
E.E..X..
........
E...E...
E..E....
........
.....E..
......EX
........
.E......
..E..EE.
5 5 100
.....
XXXXX
EEEEE
.....
XXXXX
1 1 1
E
1 1 1
P
1 1 1
X
2 2 1
E.
.X
2 2 2
E.
.X
8 1 1000
.
E
.
.
.
.
.
.
3 4 100
E...
....
...E
3 4 100
E...
.E..
...E
4 5 100
E....
.E...
...E.
.....
4 5 100
.....
.E...
.....
.....
12 12 10000000
...E.....E.E
E.E.E.....E.
.......E....
............
......E.....
............
...E.....E.E
E.E.E.....E.
.......E....
............
......E.....
............
12 12 6
...E.....E.E
E.E.E.....E.
.......E....
............
......E.....
............
...E.....E.E
E.E.E.....E.
.......E....
............
......E.....
............
12 12 10000000
E...........
............
............
............
............
............
............
............
............
............
............
............
12 12 10000000
E...........
............
............
............
............
.....E......
............
............
............
............
............
...........E
3 3 10
...
.E.
...
*/
二分匹配代码：
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
/* '0. Macros, Preprocessers, */
#pragma comment(linker, "/STACK:102400000,102400000")
#ifdef _WIN32
typedef __int64 int64;
#define OD(_TYPE) printf("%I64d
", _TYPE)
#define ODC(_CASE, _TYPE) printf("Case %d: %I64d
", _CASE++, _TYPE);
#else
typedef long long int64;
#define OD(_TYPE) printf("%lld
", _TYPE)
#define ODC(_CASE, _TYPE) printf("Case %d: %lld
", _CASE++, _TYPE);
#endif
#define LSON l,mid,id<<1
#define RSON mid+1,r,id<<1|1
#define MM (l+r)>>1
/* '1. Constants, */
const double EPS = 1e-8;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
/* '2. Coding Area, */
const int N = 15;
const int M = 205;
const int dir[4][2] = { { -1, 0 }, { 0, 1 }, { 1, 0 }, { 0, -1 } };
struct STATE {
int x, y, t;
STATE() { }
STATE(int x, int y, int t) : x(x), y(y), t(t) { }
};
int uN, vN;
vector<int> g[M];
int linker[M * M]; // 每一个v相应的u匹配
bool used[M * M];
char mp[N][N];
int dis[M][M]; // 每一个女孩到出口的最短时间
bool vis[N][N]; // 訪问过或没有
int r, c, T;
int gid, eid;
map<pair<int, int>, int> mpg, mpe; // 每一个点的编号
bool dfs(int u) {
for (unsigned int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (used[v]) continue;
used[v] = true;
if (linker[v] == -1 || dfs(linker[v])) {
linker[v] = u;
return true;
}
}
return false;
}
int hungary() {
int res = 0;
memset(linker, -1, sizeof(linker));
for (int u = 0; u < uN; u++) {
memset(used, 0, sizeof(used));
if (dfs(u))
++res;
}
return res;
}
void bfs(int x, int y) {
queue<STATE> q;
int eid = mpe[make_pair(x, y)];
memset(vis, 0, sizeof(vis));
vis[x][y] = 1;
q.push(STATE(x, y, 0));
while (!q.empty()) {
STATE now = q.front();
q.pop();
for (int i = 0; i < 4; i++) {
int dx = now.x + dir[i][0];
int dy = now.y + dir[i][1];
if (dx >= 0 && dx < r && dy >= 0 && dy < c) {
if (mp[dx][dy] != '.') continue;
if (vis[dx][dy]) continue;
int gid = mpg[make_pair(dx, dy)];
q.push(STATE(dx, dy, now.t + 1));
dis[gid][eid] = now.t + 1;
vis[dx][dy] = 1;
}
}
}
}
bool build_and_run(int limit) {
for (int i = 0; i < M; i++) g[i].clear();
uN = gid, vN = eid * limit;
for (int i = 0; i < gid; i++) {
for (int j = 0; j < eid; j++) {
for (int k = dis[i][j]; k <= limit; k++) {
g[i].push_back(j * limit + k);
}
}
}
int ans = hungary();
if (ans >= gid) return true;
return false;
}
void gao() {
gid = eid = 0;
mpg.clear(); mpe.clear();
memset(dis, 0x3f, sizeof(dis));
for (int i = 0; i < r; i++) {
scanf("%s", mp[i]);
for (int j = 0; j < c; j++) {
if (mp[i][j] == 'E') mpe[make_pair(i, j)] = eid++;
if (mp[i][j] == '.') mpg[make_pair(i, j)] = gid++;
}
}
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
if (mp[i][j] == 'E') {
bfs(i, j);
}
}
}
int L = 1, R = 256, ans = INF;
while (L <= R) {
int mid = (L + R) >> 1;
if (build_and_run(mid)) {
ans = mid;
R = mid - 1;
} else {
L = mid + 1;
}
}
if (ans == INF || ans > T) {
printf("impossible
");
} else {
printf("%d
", ans);
}
}
int main() {
while (~scanf("%d%d%d", &r, &c, &T)) {
memset(g, 0, sizeof(g));
gao();
}
return 0;
}