题意：给一个n*m的地图，'m'表示人，'H'表示房子，求所有人都回到房子所走的距离之和的最小值（距离为曼哈顿距离）。
思路：比较明显的二分图最大权匹配模型，将每个人向房子连一条边，边权为曼哈顿距离的相反数（由于是求最小，所以先取反后求最大，最后再取反回来即可），然后用KM算法跑一遍然后取反就是答案。还可以用最小费用最大流做，方法是：从源点向每个人连一条边，容量为1，费用为0，从每个房子向汇点连一条边，容量为1，费用为0，从每个人向每个房子连一条边，容量为1，费用为曼哈顿距离的值，建好图后跑一遍最小费用最大流就是答案。
附上代码：（1）KM算法，40ms左右 （2）最小费用最大流，400+ms
（1）
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/* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //#include <map> //#include <cmath> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define fillchar(a, x) memset(a, x, sizeof(a)) // //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // //template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} //template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} //template<typename T> //void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} //template<typename T> //void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} // //const double PI = acos(-1); // ///* -------------------------------------------------------------------------------- */struct KM { const static int INF = 1e9 + 7; const static int maxn = 1e3 + 7; int A[maxn], B[maxn]; int visA[maxn], visB[maxn]; int match[maxn], slack[maxn], Map[maxn][maxn]; int M, H; void add(int u, int v, int w) { Map[u][v] = w; } bool find_path ( int i ) { visA[i] = true; for ( int j = 0; j < H; j++ ) { if ( !visB[j] && A[i] + B[j] == Map[i][j] ) { visB[j] = true; if (match[j] == -1 || find_path(match[j])) { match[j] = i; return true; } } else if ( A[i] + B[j] > Map[i][j] ) //j属于B，且不在交错路径中 slack[j] = min(slack[j], A[i] + B[j] - Map[i][j]); } return false; } int solve (int M, int H) { this->M = M; this->H = H; int i, j, d; memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B)); memset(match, -1, sizeof(match)); for ( i = 0; i < M; i++ ) for ( j = 0; j < H; j++ ) A[i] = max (Map[i][j], A[i]); for ( i = 0; i < M; i++ ) { for ( j = 0; j < H; j++ ) slack[j] = INF; while ( 1 ) { memset(visA, 0, sizeof(visA)); memset(visB, 0, sizeof(visB)); if ( find_path ( i ) ) break; //从i点出发找到交错路径则跳出循环 for ( d = INF, j = 0; j < H; j++ ) //取最小的slack[j] if (!visB[j] && d > slack[j]) d = slack[j]; for ( j = 0; j < M; j++ ) //集合A中位于交错路径上的-d if ( visA[j] ) A[j] -= d; for ( j = 0; j < H; j++ ) //集合B中位于交错路径上的+d if ( visB[j] ) B[j] += d; else slack[j] -= d; //注意修改不在交错路径上的slack[j] } } int res = 0; for ( j = 0; j < H; j++ ) if (~match[j]) res += Map[match[j]][j]; return res; }};//点从0开始编号KM solver;vector<pii> H, M;int dist(pii a, pii b) { return abs(a.X - b.X) + abs(a.Y - b.Y);}int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE int n, m; while (cin >> n >> m, n || m) { H.clear(); M.clear(); for (int i = 0; i < n; i ++) { char s[123]; scanf("%s", s); for (int j = 0; s[j]; j ++) { if (s[j] == 'H') H.pb(mp(i, j)); if (s[j] == 'm') M.pb(mp(i, j)); } } for (int i = 0; i < H.size(); i ++) { for(int j = 0; j < M.size(); j ++) { solver.add(i, j, -dist(H[i], M[j])); } } cout << -solver.solve(H.size(), M.size()) << endl; } return 0;}/* ******************************************************************************** */ |
（2）
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/* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //#include <map> //#include <cmath> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define fillchar(a, x) memset(a, x, sizeof(a)) // //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // //template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} //template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} //template<typename T> //void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} //template<typename T> //void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} // //const double PI = acos(-1); // ///* -------------------------------------------------------------------------------- */struct MCMF { const static int INF = 1e9 + 7; const static int maxn = 1e5 + 7; struct Edge { int from, to, cap, cost; Edge(int u, int v, int w, int c): from(u), to(v), cap(w), cost(c) {} }; int n, s, t; vector<Edge> edges; vector<int> G[maxn]; int inq[maxn], d[maxn], p[maxn], a[maxn]; void init(int n) { this->n = n; for (int i = 0; i < n; i ++) G[i].clear(); edges.clear(); } void add(int from, int to, int cap, int cost) { edges.push_back(Edge(from, to, cap, cost)); edges.push_back(Edge(to, from, 0, -cost)); int m = edges.size(); G[from].push_back(m - 2); G[to].push_back(m - 1); } bool BellmanFord(int s, int t, int &flow, int &cost) { for (int i = 0; i < n; i ++) d[i] = INF; memset(inq, 0, sizeof(inq)); d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF; queue<int> Q; Q.push(s); while (!Q.empty()) { int u = Q.front(); Q.pop(); inq[u] = 0; for (int i = 0; i < G[u].size(); i ++) { Edge &e = edges[G[u][i]]; if (e.cap && d[e.to] > d[u] + e.cost) { d[e.to] = d[u] + e.cost; p[e.to] = G[u][i]; a[e.to] = min(a[u], e.cap); if (!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; } } } } if (d[t] == INF) return false; flow += a[t]; cost += d[t] * a[t]; int u = t; while (u != s) { edges[p[u]].cap -= a[t]; edges[p[u] ^ 1].cap += a[t]; u = edges[p[u]].from; } return true; } int solve(int s, int t) { int flow = 0, cost = 0; while (BellmanFord(s, t, flow, cost)); return cost; }};MCMF solver;vector<pii> H, M;int dist(pii a, pii b) { return abs(a.X - b.X) + abs(a.Y - b.Y);}int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE int n, m; while (cin >> n >> m, n || m) { solver.init(207); H.clear(); M.clear(); for (int i = 0; i < n; i ++) { char s[123]; scanf("%s", s); for (int j = 0; s[j]; j ++) { if (s[j] == 'H') H.pb(mp(i, j)); if (s[j] == 'm') M.pb(mp(i, j)); } } for (int i = 0; i < H.size(); i ++) solver.add(0, i + 1, 1, 0); for (int i = 0; i < M.size(); i ++) solver.add(101 + i, 201, 1, 0); for (int i = 0; i < H.size(); i ++) { for(int j = 0; j < M.size(); j ++) { solver.add(i + 1, 101 + j, 1, dist(H[i], M[j])); } } cout << solver.solve(0, 201) << endl; } return 0;}/* ******************************************************************************** */ |