题意：
给你一个n*m的地图，上面有w个人，和w个房子，每个人都要进房子，每个房子只能进一个人，问所有人都进房子的路径总和最少是多少？
思路：
比较简单的最大流，直接建立两排，左边人，右边房子，广搜或者深搜求距离建图，然后一边费用流就行了，比较简单，没啥说的地方，就这样。
#include<queue>
#include<stdio.h>
#include<string.h>
#define N_node 200 + 10
#define N_edge (200 * 200 + 200 ) * 2 + 100
#define INF 1000000000
using namespace std;
typedef struct
{
int from ,to ,cost ,flow ,next;
}STAR;
typedef struct
{
int x ,y ,t;
}NODE;
STAR E[N_edge];
NODE xin ,tou;
int list[N_node] ,tot;
int s_x[N_node];
int mer[N_node];
int map[102][102] ,n ,m;
int dir[4][2] = {0 ,1 ,0 ,-1 ,1 ,0 ,-1 ,0};
void add(int a ,int b ,int c ,int d)
{
E[++tot].from = a;
E[tot].to = b;
E[tot].cost = c;
E[tot].flow = d;
E[tot].next = list[a];
list[a] = tot;
E[++tot].from = b;
E[tot].to = a;
E[tot].cost = -c;
E[tot].flow = 0;
E[tot].next = list[b];
list[b] = tot;
}
bool SPFA(int s ,int t ,int n)
{
for(int i = 0 ;i <= n ;i ++)
s_x[i] = INF;
bool mark[N_node] = {0};
queue<int>q;
q.push(s);
mark[s] = 1;
s_x[s] = 0;
memset(mer ,255 ,sizeof(mer));
while(!q.empty())
{
int xin ,tou;
tou = q.front();
q.pop();
mark[tou] = 0;
for(int k = list[tou] ;k ;k = E[k].next)
{
xin = E[k].to;
if(s_x[xin] > s_x[tou] + E[k].cost && E[k].flow)
{
s_x[xin] = s_x[tou] + E[k].cost;
mer[xin] = k;
if(!mark[xin])
{
mark[xin] = 1;
q.push(xin);
}
}
}
}
return mer[t] != -1;
}
int M_C_Flow(int s ,int t ,int n)
{
int mincost = 0 ,maxflow = 0 ,minflow;
while(SPFA(s ,t ,n))
{
minflow = INF;
for(int i = mer[t] ;i + 1 ;i = mer[E[i].from])
if(minflow > E[i].flow) minflow = E[i].flow;
for(int i = mer[t] ;i + 1 ;i = mer[E[i].from])
{
E[i].flow -= minflow;
E[i^1].flow += minflow;
mincost += E[i].cost;
}
maxflow += minflow;
}
return mincost;
}
void BFS_Buid(int x ,int y ,int N)
{
xin.x = x ,xin.y = y ,xin.t = 0;
int mark[105][105] = {0};
queue<NODE>q;
mark[x][y] = 1;
q.push(xin);
while(!q.empty())
{
tou = q.front();
q.pop();
if(map[tou.x][tou.y] > N)
add(map[x][y] ,map[tou.x][tou.y] ,tou.t ,INF);
for(int i = 0 ;i < 4 ;i ++)
{
xin.x = tou.x + dir[i][0];
xin.y = tou.y + dir[i][1];
xin.t = tou.t + 1;
if(xin.x >= 1 && xin.x <= n && xin.y >= 1 && xin.y <= m && !mark[xin.x][xin.y])
{
mark[xin.x][xin.y] = 1;
q.push(xin);
}
}
}
return ;
}
int main ()
{
int i ,j;
char str[105][105];
while(~scanf("%d %d" ,&n ,&m) && n + m)
{
int M = 0 ,H = 0 ,N = 0;
for(i = 1 ;i <= n ;i ++)
{
scanf("%s" ,str[i]);
for(j = 0 ;j < m ;j ++)
if(str[i][j] == 'm')
N ++;
}
for(i = 1 ;i <= n ;i ++)
{
for(j = 1 ;j <= m ;j ++)
{
if(str[i][j-1] == '.') map[i][j] = 0;
else if(str[i][j-1] == 'm') map[i][j] = ++M;
else
{
++H;
map[i][j] = H + N;
}
}
}
memset(list ,0 ,sizeof(list)) ,tot = 1;
for(i = 1 ;i <= N ;i ++)
add(0 ,i ,0 ,1) ,add(i + N ,N + N + 1 ,0 ,1);
for(i = 1 ;i <= n ;i ++)
for(j = 1 ;j <= m ;j ++)
{
if(map[i][j] && map[i][j] <= N)
BFS_Buid(i ,j ,N);
}
printf("%d
" ,M_C_Flow(0 ,N + N + 1 ,N + N + 1));
}
return 0;
}