题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1189
题目大意:给你一个n*m的区域,有门和墙,每块空地上有一个人,疏散开始后每块空地可以站无数个人,但是门每次只能经过一个人,请问最少需要多少时间才能全部撤到门外
建一个源点s连接到所有空地,流量为1,然后二分最长时间,空地向可以到达的门连一条边(这里的可到达是指在二分出的时间内可到达),所有门向汇点t连一条容量为时间的边,跑最大流,如果满流,就降低上限,否则提高下限。(似乎直接从1暴力跑下去也不会TLE……)
#include
#include
#include
#include
#include
#include
#define INF 214748364
using namespace std;
struct Link
{
int s,t,c,next;
}l[100000];
Link l2[100000];
int g[5000];
int cnt[5000];
int dist[5000];
bool b[5000];
char a[30][30];
int d[30][30];
queue<int> Q;
int n,m,p,sum = 0;
bool bfs(int x,int y)
{
Q.push(x);
dist[x] = 1;
while(!Q.empty())
{
int q = Q.front();
int w = g[q];
while(w)
{
if(l[w].c && !dist[l[w].t])
{
dist[l[w].t] = dist[q] + 1;
Q.push(l[w].t);
}
w = l[w].next;
}
Q.pop();
}
return dist[y];
}
int dfs(int x,int t,int flow)
{
if(x == t || flow == 0)
return flow;
int w = cnt[x];
while(w)
{
if(dist[l[w].t] == dist[x] + 1 && !b[l[w].t] && l[w].c)
{
b[l[w].t] = 1;
int k = dfs(l[w].t,t,min(flow,l[w].c));
b[l[w].t] = 0;
if(k)
{
cnt[x] = w;
l[w].c -= k;
l[w^1].c += k;
return k;
}
}
w = l[w].next;
}
return 0;
}
int maxflow(int s,int t)
{
memset(dist,0,sizeof(dist));
int flow = 0;
while(bfs(s,t))
{
while(233)
{
int w = dfs(s,t,INF);
memset(b,0,sizeof(b));
flow += w;
if(w == 0)
break;
}
for(int i = 0;i <= t;i ++)
cnt[i] = g[i];
memset(dist,0,sizeof(dist));
}
return flow;
}
void Add_Link(int s,int t,int c,int x)
{
l[x].s = s;
l2[x].s = l[x].s;
l[x].t = t;
l2[x].t = l[x].t;
l[x].c = c;
l2[x].c = l[x].c;
l[x].next = g[s];
g[s] = x;
l[x^1].s = t;
l2[x^1].s = l[x^1].s;
l[x^1].t = s;
l2[x^1].t = l[x^1].t;
l[x^1].c = 0;
l2[x^1].c = l[x^1].c;
l[x^1].next = g[t];
g[t] = x^1;
return ;
}
void bbffss(int x,int y,int time)
{
int k = (x-1)*(m+1) + y;
Q.push(k);
while(!Q.empty())
{
int e = Q.front();
x = e / (m+1) + 1;
y = e % (m+1);
p += 2;
Add_Link((x-1)*(m+1)+y,k,1,p);
if(!d[x+1][y] && a[x+1][y] == '.' )
{
d[x+1][y] = d[x][y] + 1;
if(d[x+1][y] <= time)
Q.push((x)*(m+1)+y);
}
if(!d[x-1][y] && a[x-1][y] == '.')
{
d[x-1][y] = d[x][y] + 1;
if(d[x-1][y] <= time)
Q.push((x-2)*(m+1)+y);
}
if(!d[x][y+1] && a[x][y+1] == '.')
{
d[x][y+1] = d[x][y] + 1;
if(d[x][y+1] <= time)
Q.push((x-1)*(m+1)+y+1);
}
if(!d[x][y-1] && a[x][y-1] == '.')
{
d[x][y-1] = d[x][y] + 1;
if(d[x+1][y] <= time)
Q.push((x-1)*(m+1)+y-1);
}
Q.pop();
}
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= m;j ++)
{
d[i][j] = 0;
}
}
return ;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;i ++)
{
cin >> a[i];
for(int j = m;j ;j --)
{
a[i][j] = a[i][j-1];
if(a[i][j] == '.')
{
p += 2;
Add_Link(0,(i-1)*(m+1)+j,1,p);
sum ++;
}
}
a[i][0] = 'X';
a[i][m+1] = 'X';
}
for(int i = 1;i <= m;i ++)
{
a[0][i] = 'X';
a[n+1][i] = 'X';
}
int s = 1,t = 1000,k = p + 2;
while(s != t)
{
int mid = (s+t)/2;
p = 0;
memset(l,0,sizeof(l));
memset(g,0,sizeof(g));
for(p ;p < k;p += 2)
Add_Link(l2[p].s,l2[p].t,l2[p].c,p);
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= m;j ++)
{
if(a[i][j] == 'D')
{
bbffss(i,j,mid);
p += 2;
Add_Link((i-1)*(m+1)+j,4999,mid,p);
}
}
}
if(maxflow(0,4999) == sum)
t = mid;
else
s = mid + 1;
}
if(s == 1000)
{
cout << "impossible" << endl;
return 0;
}
cout << s << endl;
return 0;
}