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\n", -gao.km());
}
return 0;
}