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.
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
题意没什么好说的 看样例就知道题意了
直接套板子
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <cmath> 5 #include <algorithm> 6 #include <set> 7 #include <iostream> 8 #include <map> 9 #include <stack> 10 #include <string> 11 #include <vector> 12 #define pi acos(-1.0) 13 #define eps 1e-6 14 #define fi first 15 #define se second 16 #define lson l,m,rt<<1 17 #define rson m+1,r,rt<<1|1 18 #define bug printf("******\n") 19 #define mem(a,b) memset(a,b,sizeof(a)) 20 #define fuck(x) cout<<"["<<x<<"]"<<endl 21 #define f(a) a*a 22 #define sf(n) scanf("%d", &n) 23 #define sff(a,b) scanf("%d %d", &a, &b) 24 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c) 25 #define pf printf 26 #define FRE(i,a,b) for(i = a; i <= b; i++) 27 #define FREE(i,a,b) for(i = a; i >= b; i--) 28 #define FRL(i,a,b) for(i = a; i < b; i++) 29 #define FRLL(i,a,b) for(i = a; i > b; i--) 30 #define FIN freopen("DATA.txt","r",stdin) 31 #define lowbit(x) x&-x 32 #pragma comment (linker,"/STACK:102400000,102400000") 33 34 using namespace std; 35 const int maxn = 1e5 + 10; 36 typedef long long LL; 37 const int MX = 505; 38 const int inf = 0x3f3f3f3f; 39 const int MXE = MX * MX * 4; 40 struct MinCost_MaxFlow { 41 struct Edge { 42 int v, w, nxt; 43 int cost; 44 } E[MXE]; 45 int head[MX], tot, level[MX], pre[MX], d[MX]; 46 bool vis[MX]; 47 void init() { 48 memset(head, -1, sizeof(head)); 49 tot = 0; 50 } 51 void add(int u, int v, int w, int cost) { 52 E[tot].v = v; 53 E[tot].w = w; 54 E[tot].cost = cost; 55 E[tot].nxt = head[u]; 56 head[u] = tot++; 57 E[tot].v = u; 58 E[tot].w = 0; 59 E[tot].cost = -cost; 60 E[tot].nxt = head[v]; 61 head[v] = tot++; 62 } 63 bool spfa(int s, int t) { 64 memset(vis, 0, sizeof(vis)); 65 memset(d, 0x3f, sizeof(d)); 66 memset(pre, -1, sizeof(pre)); 67 queue<int>q; 68 q.push(s); 69 d[s] = 0; 70 vis[s] = 1; 71 while (!q.empty()) { 72 int u = q.front(); 73 q.pop(); 74 vis[u] = 0; 75 for (int i = head[u]; ~i; i = E[i].nxt) { 76 int w = E[i].w, v = E[i].v, cost = E[i].cost; 77 if (w > 0 && d[v] > d[u] + cost) { 78 d[v] = d[u] + cost; 79 pre[v] = i; 80 if (!vis[v]) { 81 q.push(v); 82 vis[v] = 1; 83 } 84 } 85 } 86 } 87 //如果是最小费用可行流则要这一句(要求费用最小,不要求流量最大) 88 //if (d[t] > 0) return false; 89 return pre[t] != -1; 90 } 91 int solve(int s, int t, int &cost) { 92 int flow = 0; 93 cost = 0; 94 while (spfa(s, t)) { 95 int minFlow = inf; 96 for (int i = pre[t]; ~i; i = pre[E[i ^ 1].v]) 97 minFlow = min(minFlow, E[i].w); 98 for (int i = pre[t]; ~i; i = pre[E[i ^ 1].v]) { 99 cost += minFlow * E[i].cost; 100 E[i].w -= minFlow; 101 E[i ^ 1].w += minFlow; 102 } 103 flow += minFlow; 104 } 105 return flow; 106 } 107 } F; 108 int n, m; 109 struct Point { 110 int x, y; 111 Point (int x, int y): x(x), y(y) {} 112 }; 113 int cal(Point a, Point b) { 114 return abs(a.x - b.x) + abs(a.y - b.y); 115 } 116 char tu[105][105]; 117 vector<Point>men; 118 vector<Point>home; 119 int main() { 120 while(~sff(n, m), n + m) { 121 F.init(); 122 men.clear(); 123 home.clear(); 124 for (int i = 0 ; i < n ; i++) { 125 scanf("%s", tu[i]); 126 for (int j = 0 ; j < m ; j++) { 127 if (tu[i][j] == 'm') men.push_back(Point(i, j)); 128 if (tu[i][j] == 'H') home.push_back(Point(i, j)); 129 } 130 } 131 int s=0,len1=men.size(),len2=home.size(),t; 132 t=len1+len2+1; 133 for (int i=0 ;i<len1 ;i++) 134 for (int j=0 ;j<len2 ;j++) 135 F.add(i+1,j+1+len1,1,cal(men[i],home[j])); 136 for (int i=1 ;i<=len1 ;i++) F.add(0,i,1,0); 137 for (int i=1 ;i<=len2 ;i++) F.add(i+len1,t,1,0); 138 int cost = 0; 139 F.solve(s, t, cost); 140 printf("%d\n", cost); 141 } 142 return 0; 143 }