Poj2195going home [km]

Question:

Given a matrix, M representative h Represents the house. Now I want to move the minimum number of steps back to the house.

 

Analysis:

You can use a bipartite graph or a network stream to write data.

Bipartite Graph: X sets people's y sets as houses, and each house's connected edge is a-Dist edge. You can find the maximum weight, and then obtain the inverse value.

Network Flow: Use the billing flow. The building capacity between people and houses is 1, and the cost is Dist. You can find the minimum cost.

 

Because there are edges between people and every house, I use the adjacent matrix to process

Bipartite Graph code:

  1 #include <iostream>  2 #include <cstdio>  3 #include <cstring>  4 #include <map>  5 #include <vector>  6 #include <cmath>  7 using namespace std;  8   9 const int maxn = 105; 10 const int INF = 1000000000; 11  12 typedef pair<int, int> PII; 13 vector<PII> v[2]; 14  15  16 int n_cnt; 17 int W[maxn][maxn]; 18 int Lx[maxn], Ly[maxn]; 19 int Left[maxn]; 20 bool S[maxn], T[maxn]; 21  22 bool match(int i) { 23     S[i] = true; 24     for(int j = 1; j <= n_cnt; j++) if(Lx[i] + Ly[j] == W[i][j] && !T[j]) { 25         T[j] = true; 26         if(!Left[j] || match(Left[j])) { 27             Left[j] = i; 28             return true; 29         } 30     } 31     return false; 32 } 33  34  35 void update() { 36     int a = INF; 37     for(int i = 1; i <= n_cnt; i++) if(S[i]) { 38         for(int j = 1; j <= n_cnt; j++) if(!T[j]) { 39             a = min(a, Lx[i] + Ly[j] - W[i][j]); 40         } 41     } 42     for(int i = 1; i <= n_cnt; i++) { 43         if(S[i]) Lx[i] -= a; 44         if(T[i]) Ly[i] += a; 45     } 46 } 47  48 void KM() { 49     for(int i = 1; i <= n_cnt; i++) { 50         Left[i] = Lx[i] = Ly[i] = 0; 51         for(int j = 1; j <= n_cnt; j++) { 52             Lx[i] = max(Lx[i], W[i][j]); 53         } 54     } 55     for(int i = 1; i <= n_cnt; i++) { 56         for(;;) { 57             for(int j = 1; j <= n_cnt; j++) S[j] = T[j] = 0; 58             if(match(i)) break; 59             else update(); 60         } 61     } 62 } 63  64 int dist(PII p1, PII p2) { 65     return fabs(p1.first - p2.first) + fabs(p1.second - p2.second); 66 } 67  68 int main() { 69     int n, m; 70 //    freopen("a.txt","r",stdin); 71     while(scanf("%d %d",&n, &m) && n + m) { 72         char mat[maxn][maxn]; 73         memset(W, 0, sizeof(W)); 74         v[0].clear(); v[1].clear(); 75         for(int i = 0; i < n; i++) { 76                 scanf("\n%s",mat[i]); 77                // puts(mat[i]); 78         } 79         for(int i = 0; i < n; i++) { 80             for(int j = 0; j < m; j++) { 81                 if(mat[i][j] == ‘m‘) 82                     v[0].push_back(make_pair(i, j)); 83                 else if(mat[i][j] == ‘H‘) 84                     v[1].push_back(make_pair(i, j)); 85             } 86         } 87         n_cnt = v[0].size(); 88 //        printf("n == %d\n",n_cnt); 89         for(int i = 0; i < n_cnt; i++) { 90             for(int j = 0; j < n_cnt; j++) { 91                 W[i + 1][j + 1] = -dist(v[0][i], v[1][j]); 92             } 93         } 94         KM(); 95         int ans = 0; 96         for(int i = 1; i <= n_cnt; i++) { 97             ans += Lx[i] + Ly[i]; 98         } 99         printf("%d\n",-ans);100     }101     return 0;102 }

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