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.
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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.
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
題目大意：圖中有n個man和n個home，並且一個人只能住在一個房子裏面，房子和人的個數是相等的。並且每個人移動一步的代價是1，怎麼使
所有人住在房子裏，並且使所有人的代價和最小。
給出一個n×m的圖，m表示人，H表示房子，.表示空地，當然房子不算障礙物，可以穿過
解題思路：這是一個費用流的應用，構圖如下：
建一個源點指向所有人，容量爲1，代價爲0
使每個人指向所有的房子，容量爲1，代價爲人與房子的曼哈頓距離
建一個匯點使所有房子指向它，容量爲1，代價爲0
然後從源點到匯點求費用流即可。
題目要注意的是，在n×m的這張圖上說人的總數不超過100，那麼圖中節點總數爲人+房子+源點+匯點=202
這個題不知道怎麼回事，我做的老是不對，真是奇了怪了，數據沒錯啊！
有誰給我看看一下的代碼怎麼錯了嗎？不勝感激
代碼（我的錯誤的）
#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;
}
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;
}代碼菜鳥，如有錯誤，請多包涵！！！
如有幫助記得支持我一下，謝謝！！！