Going Home poj 2195

Going Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22062		Accepted: 11144

Description

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters
a house. The task is complicated with the restriction that each house can accommodate only one little man. 

Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a '.' means an empty space, an 'H' represents a house on that point, and am 'm' indicates
there is a little man on that point. 



You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.
Input

There are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both
N and M are between 2 and 100, inclusive. There will be the same number of 'H's and 'm's on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.
Output

For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.
Sample Input

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

Sample Output

2
10
28

题目大意：图中有n个man和n个home，并且一个人只能住在一个房子里面，房子和人的个数是相等的。并且每个人移动一步的代价是1，怎么使

所有人住在房子里，并且使所有人的代价和最小。

给出一个n×m的图，m表示人，H表示房子，.表示空地，当然房子不算障碍物，可以穿过

解题思路：这是一个费用流的应用，构图如下：

建一个源点指向所有人，容量为1，代价为0

使每个人指向所有的房子，容量为1，代价为人与房子的曼哈顿距离

建一个汇点使所有房子指向它，容量为1，代价为0

然后从源点到汇点求费用流即可。

题目要注意的是，在n×m的这张图上说人的总数不超过100，那么图中节点总数为人+房子+源点+汇点=202

这个题不知道怎么回事，我做的老是不对，真是奇了怪了，数据没错啊！

有谁给我看看一下的代码怎么错了吗？不胜感激

代码（我的错误的）

#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
using namespace std;

const int inf = INT_MAX;
struct node
{
int from, to, cap, flow;
int cost, next;
}E[100002];
int sum, top;
int head[1100], pre[1100];
int dist[1100];
bool vis[1100];
char ch[10100];

void inti()
{
top = 0;
memset ( head, -1, sizeof(head) );
}

void AddEdge(int u, int v, int w, int c)
{
node temp = {u, v, w, 0, c, head[u]};
E[top] = temp;
head[u] = top++;
node temp1 = {v, u, 0, 0, -c, head[v]};
E[top] = temp1;
head[v] = top++;
}

int juli(int x, int y, int m)
{
return abs(y/m-x/m)+abs(y%m-x%m);
}

bool SPFA()
{
int i;
queue<int> q;
for ( i = 0;i <= sum+6; i++ )
dist[i] = inf;
memset ( vis, 0, sizeof(vis) );
memset( pre, -1, sizeof(pre) );
dist[0] = 0;
vis[0] = true;
q.push(0);
while ( !q.empty() )
{
int u = q.front();
q.pop();
vis[u] = false;
for ( i = head[u];i != -1; i = E[i].next )
{
node temp = E[i];
if ( dist[temp.to] > dist[u]+temp.cost&&temp.cap > temp.flow )
{
dist[temp.to] = dist[u]+temp.cost;
pre[temp.to] = i;
if ( !vis[temp.to] )
{
vis[temp.to] = true;
q.push(temp.to);
}
}
}
}
return dist[sum] != inf;
}

int main()
{
int n, m, i, j;
while ( ~scanf ( "%d %d%*c", &n, &m )&&(n || m) )
{
inti();
int hm[1001], mm[1001];
int sum1 = 0, sum2 = 0;
for ( i = 0;i < n; i++ )
{
scanf ( "%s", ch );
for ( j = 0;j < m; j++ )
{
if ( ch[j] == 'm' )
mm[++sum1] = i*m+j;
else if ( ch[j] == 'H' )
hm[++sum2] = i*m+j;
}
getchar();
}
sum = sum1+sum2;
for ( i = 1;i <= sum1; i++ )
AddEdge(0, i, 1, 0);
for ( i = 1;i <= sum2; i++ )
AddEdge(i+sum1, sum, 1, 0);
for ( i = 1;i <= sum1; i++ )
{
for ( j = 1;j <= sum2; j++ )
{
int ju = juli(mm[i], hm[j], m);
AddEdge(i, j+sum1, 1, ju);
}
}
int cost = 0;
while ( SPFA() )
{
int Min = inf;
for( i = pre[sum]; i != -1; i = pre[E[i^1].to])
{
node temp = E[i];
Min = min(Min, temp.cap-temp.flow);
}
for( i = pre[sum]; i != -1; i = pre[E[i^1].to])
{
E[i].flow = E[i].flow+Min;
E[i^1].flow = E[i^1].flow-Min;
cost = cost+E[i].cost;
}
}
printf ( "%d\n", cost );
}
return 0;
}
代码（正确的）
Memory: 5532K Time: 125MS
Language: C++ Result: Accepted
Source Code
#include <cstring>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <iostream>
using namespace std;

const int inf = INT_MAX;
struct node
{
int u, v;
int next;
}E[1000002];
int num, nos, pre[1005];
int head[1005], cost[1005][1005];

void AddEdge(int l, int r, int v)
{
E[num].u = r;
E[num].v = v;
E[num].next = head[l];
head[l] = num++;
}

int juli(int x, int y, int m)
{
return abs(y/m-x/m)+abs(y%m-x%m);
}

int bfs()
{
int visit[1001];
int dist[1001];
int i;
for ( i = 0;i <= nos; i++ )
dist[i] = inf;
memset ( visit, 0, sizeof(visit) );
memset( pre, -1, sizeof(pre) );
queue<int> q;
q.push(0);
visit[0] = 1;
dist[0] = 0;
while ( !q.empty() )
{
int e = q.front();
q.pop();
visit[e] = 0;
for ( i = head[e];i != -1; i = E[i].next )
{
int r = E[i].u;
if ( dist[e]+E[i].v < dist[r] && cost[e][r] )
{
pre[r] = e;
dist[r] = dist[e]+E[i].v;
if ( !visit[r] )
{
q.push(r);
visit[r] = 1;
}
}
}
}
if ( dist[nos] != inf )
return 1;
return 0;
}

void change()
{
int minx = inf;
int i;
for ( i = nos;pre[i] != -1; i = pre[i] )
minx = min(minx, cost[pre[i]][i]);
for ( i = nos;pre[i] != -1; i = pre[i] )
{
cost[pre[i]][i] -= minx;
cost[i][pre[i]] += minx;
}
}

int main()
{
int n, m, i, j;
char str[100001];
while ( scanf("%d %d",&n,&m)&&(n||m) )
{
memset ( head, -1, sizeof(head) );
memset ( cost, 0, sizeof(cost) );
int hm, mm;
hm = mm = 0;
int ms[100001];
int hs[100001];
for ( i = 0;i < n; i++ )
{
scanf ( "%s", str );
for ( j = 0;j < m; j++ )
{
if ( str[j] == 'm' )
ms[++mm] = i*m+j;
else if ( str[j] == 'H' )
hs[++hm] = i*m+j;
}
}
nos = hm+mm+1;
for ( i = 1;i <= mm; i++ )
{
AddEdge(0, i, 0);
AddEdge(i, 0, 0);
cost[0][i] = 1;
cost[i][0] = 0;
}
for ( j = 1;j <= hm; j++ )
{
AddEdge(j+mm, mm+hm+1, 0);
AddEdge(mm+hm+1, j+mm, 0);
cost[j+mm][mm+hm+1] = 1;
cost[mm+hm+1][j+mm] = 0;
}
for ( i = 1;i <= mm; i++ )
{
for ( j = 1;j <= hm; j++ )
{
int ju = juli(ms[i], hs[j], m);
AddEdge(i, j+mm, ju);
AddEdge(j+mm, i, -ju);
cost[i][j+mm] = inf;
cost[j+mm][i] = 0;
}
}
while ( bfs() )
change();
int sum = 0;
for ( i = 1;i <= mm; i++ )
{
for ( j = 1;j <= hm; j++ )
{
sum += (inf-cost[i][j+mm])*juli(ms[i], hs[j], m);
}
}
printf ( "%d\n", sum );
}
return 0;
}

代码菜鸟，如有错误，请多包涵！！！
如有帮助记得支持我一下，谢谢！！！