本文主要是介绍HDU 1533 Going Home（KM完美匹配），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
HDU 1533 Going Home
题目链接
题意：就是一个H要对应一个m，使得总曼哈顿距离最小
思路：KM完美匹配，由于是要最小，所以边权建负数来处理即可
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\n", -gao.km());}return 0;
}这篇关于HDU 1533 Going Home（KM完美匹配）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
