Going Home
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 17230 | Accepted: 8781 |
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
Source
题意：
5 5
HH..m
.....
.....
.....
mm..H
对于一个图
m代表人H代表房间
现在需要用最少的步数将所有的人送回房间，问最小步数是多少，每个房间只能容纳一个人
分析：
每个人到每个房间的花费我们可以求出来
现在，我们把人和花费都抽象成一个一个的点，将人和房间连一条花费为步数的边
那么求出最小费用最大流即可
代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <vector>
5 #include <cmath>
6 #include <queue>
7 using namespace std;
8
9 const int maxn = 1005;
10
11 typedef pair<int, int> PII;
12
13 char a[maxn][maxn];
14 vector<PII>v[2];
15 const int INF = 1000000000;
16
17 struct Edge {
18 int from, to, cap, flow, cost;
19 };
20
21 struct MCMF
22 {
23 int n, m, s, t;
24 vector<Edge> edges;
25 vector<int> G[maxn];
26
27 int inq[maxn];
28 int d[maxn];
29 int p[maxn];
30 int a[maxn];
31
32 void init(int n) {
33 this -> n = n;
34 for(int i = 0; i < n; i++) G[i].clear();
35 edges.clear();
36 }
37
38 void AddEdge(int from, int to, int cap, int cost) {
39 edges.push_back((Edge) { from, to, cap, 0, cost } );
40 edges.push_back((Edge) { to, from, 0, 0, -cost } );
41 m = edges.size();
42 G[from].push_back(m - 2);
43 G[to].push_back(m - 1);
44 }
45
46 bool BellmanFord(int s, int t, int &flow, int &cost) {
47 for(int i = 0; i < n; i++) d[i] = INF;
48 memset(inq, 0, sizeof(inq) );
49 d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
50 queue<int> Q;
51 Q.push(s);
52 while(!Q.empty()) {
53 int u = Q.front(); Q.pop();
54 inq[u] = 0;
55 for(int i = 0; i < G[u].size(); i++) {
56 Edge &e = edges[G[u][i]];
57 if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
58 d[e.to] = d[u] + e.cost;
59 p[e.to] = G[u][i];
60 a[e.to] = min(a[u], e.cap - e.flow);
61 if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
62 }
63 }
64 }
65
66 if(d[t] == INF) return false;
67 flow += a[t];
68 cost += d[t] * a[t];
69 int u = t;
70 while(u != s) {
71 edges[p[u]].flow += a[t];
72 edges[p[u] ^ 1].flow -= a[t];
73 u = edges[p[u]].from;
74 }
75 return true;
76 }
77
78 int MinCost(int s, int t) {
79 //this -> s = s; this -> t = t;
80 int flow = 0, cost = 0;
81 while(BellmanFord(s, t, flow, cost)){};
82 return cost;
83 }
84 };
85
86 MCMF g;
87
88 int get_dist(PII p1, PII p2) {
89 return fabs(p1.first - p2.first) + fabs(p1.second - p2.second);
90 }
91
92 int main() {
93 int n, m;
94 //freopen("a.txt","r",stdin);
95 while(scanf("%d %d",&n, &m)) {
96 if(n == 0 && m == 0)
97 break;
98 //puts("*");
99 int m_num = 0, H_num = 0;
100 for(int i = 0; i < 2; i++) {
101 v[i].clear();
102 }
103 g.init(n * m);//节点个数 尽量大一点
104 for(int i = 0; i < n; i++) {
105 for(int j = 0; j < m; j++) {
106 scanf("\n%c",&a[i][j]);//
107 if(a[i][j] == ‘m‘) {
108 m_num++;
109 v[0].push_back(make_pair(i, j));//用vector把所有的人和房间都储存下来
110 }
111 else if(a[i][j] == ‘H‘) {
112 H_num++;
113 v[1].push_back(make_pair(i, j));
114 }
115 }
116 }
117 int s = 0; int t = m_num + H_num + 1;
118 //printf("%d %d %d %d\n",m_num, H_num, v[0].size(), v[1].size());
119 for(int i = 1; i <= m_num; i++) {
120 g.AddEdge(s, i, 1, 0);//源点跟人的容量为1 花费为 0 因为 只能输送一个人
121 }
122 for(int i = m_num + 1; i <= m_num + H_num; i++) {
123 g.AddEdge(i, t, 1, 0);//同理
124 }
125
126 for(int i = 0; i < v[0].size(); i++) {
127 for(int j = 0; j < v[1].size(); j++) {
128 g.AddEdge(i + 1, m_num + j + 1, 1, get_dist(v[0][i], v[1][j]));//容量为1 花费为步数
129 }
130 }
131 printf("%d\n",g.MinCost(s, t)) ;
132 }
133 return 0;
134 }
hdu2195Going Home【最小费用最大流】,布布扣,bubuko.com