POJ3020《Antenna Placement》方法：无向图最小路径覆盖 匈牙利算法

就是要求无向图最小路径覆盖问题，可以将每个点视为2个拆点i,i'，从而分为两个顶点集，利用 最小路径覆盖数 = 顶点数 - 最大二分匹配数/2求得。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int city[41][11]; //第num个城市的坐标bool map[401][401]; //第i个和第j个城市间可以连通int num; //城市数bool v[401];int link[401]; // 匹配哪个V1中的点int match;int row[4] = {0, 1, 0,-1};int col[4] = {1, 0,-1, 0};bool dfs(int x){for (int y = 1; y <= num; ++y) if (map[x][y] && !v[y]) {v[y] = true;if (link[y] == 0 || dfs(link[y])) {link[y] = x;return true;}}return false;}void hungary(){for (int i = 1; i <= num; ++i) {memset(v, 0, sizeof(v));if (dfs(i))++match;}}int main(){//freopen("temp.txt", "r", stdin);int test, h, w, x, y;char c;cin >> test;while (test--) {memset(city, 0, sizeof(city));memset(map, 0, sizeof(map));memset(link, 0, sizeof(link));num = match = 0;cin >> h >> w; //行数，列数for (int i = 1; i <= h; ++i)for (int j = 1; j <= w; ++j) {cin >> c;if (c == '*')city[i][j] = ++num;}for (int i = 1; i <= h; ++i) {for (int j = 1; j <= w; ++j) if (city[i][j]) for (int k = 0; k <= 3; ++k) {x = i + row[k];y = j + col[k];if (city[x][y])map[ city[i][j] ][ city[x][y] ] = true;}}hungary();cout << num-match/2<< endl;}return 0;}