HDU 1533 Going Home
题意:就是一个H要相应一个m,使得总曼哈顿距离最小
思路:KM完美匹配,因为是要最小。所以边权建负数来处理就可以
代码:
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int MAXNODE = 105; typedef int Type; const Type INF = 0x3f3f3f3f; struct KM { int n; Type g[MAXNODE][MAXNODE]; Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE]; int left[MAXNODE]; bool S[MAXNODE], T[MAXNODE]; void init(int n) { this->n = n; } void add_Edge(int u, int v, Type val) { g[u][v] = val; } bool dfs(int i) { S[i] = true; for (int j = 0; j < n; j++) { if (T[j]) continue; Type tmp = Lx[i] + Ly[j] - g[i][j]; if (!tmp) { T[j] = true; if (left[j] == -1 || dfs(left[j])) { left[j] = i; return true; } } else slack[j] = min(slack[j], tmp); } return false; } void update() { Type a = INF; for (int i = 0; i < n; i++) if (!T[i]) a = min(a, slack[i]); for (int i = 0; i < n; i++) { if (S[i]) Lx[i] -= a; if (T[i]) Ly[i] += a; } } int km() { for (int i = 0; i < n; i++) { left[i] = -1; Lx[i] = -INF; Ly[i] = 0; for (int j = 0; j < n; j++) Lx[i] = max(Lx[i], g[i][j]); } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) slack[j] = INF; while (1) { for (int j = 0; j < n; j++) S[j] = T[j] = false; if (dfs(i)) break; else update(); } } int ans = 0; for (int i = 0; i < n; i++) ans += g[left[i]][i]; return ans; } } gao; const int N = 105; int n, m; char str[N]; struct Point { int x, y; Point() {} Point(int x, int y) { this->x = x; this->y = y; } } hp[N], mp[N]; int dis(Point a, Point b) { return abs(a.x - b.x) + abs(a.y - b.y); } int hn, mn; int main() { while (~scanf("%d%d", &n, &m) && n || m) { hn = mn = 0; for (int i = 0; i < n; i++) { scanf("%s", str); for (int j = 0; j < m; j++) { if (str[j] == 'H') hp[hn++] = Point(i, j); if (str[j] == 'm') mp[mn++] = Point(i, j); } } gao.n = hn; for (int i = 0; i < hn; i++) { for (int j = 0; j < mn; j++) { gao.g[i][j] = -dis(hp[i], mp[j]); } } printf("%d ", -gao.km()); } return 0; }