HDU 3523 Image copy detection（KM最大匹配）

HDU 3523 Image copy detection

题目链接

题意：这题事实上题意读懂就简单了，说白了就是1-n放到1-n列，每列的值为每列上数字和该数字的差的绝对值，然后求总和最小

思路：就一KM最大匹配

代码：

#include <cstdio>

#include <cstring>

#include <cmath>

#include <cstdlib>

#include <algorithm>

using namespace std;

const int MAXNODE = 505;

typedef int Type;

const Type INF = 0x3f3f3f3f;

struct KM {

	int n, m;

	Type g[MAXNODE][MAXNODE];

	Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];

	int left[MAXNODE], right[MAXNODE];

	bool S[MAXNODE], T[MAXNODE];

	void init(int n, int m) {

		this->n = n;

		this->m = m;

		memset(g, 0, sizeof(g));

	}

	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 < m; 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;

					right[i] = j;

					return true;

				}

			} else slack[j] = min(slack[j], tmp);

		}

		return false;

	}

	void update() {

		Type a = INF;

		for (int i = 0; i < m; i++)

			if (!T[i]) a = min(a, slack[i]);

		for (int i = 0; i < n; i++)

			if (S[i]) Lx[i] -= a;

		for (int i = 0; i < m; i++)

			if (T[i]) Ly[i] += a;

	}

	Type km() {

		memset(left, -1, sizeof(left));

		memset(right, -1, sizeof(right));

		memset(Ly, 0, sizeof(Ly));

		for (int i = 0; i < n; i++) {

			Lx[i] = -INF;

			for (int j = 0; j < m; j++)

				Lx[i] = max(Lx[i], g[i][j]);

		}

		for (int i = 0; i < n; i++) {

			for (int j = 0; j < m; j++) slack[j] = INF;

			while (1) {

				memset(S, false, sizeof(S));

				memset(T, false, sizeof(T));

				if (dfs(i)) break;

				else update();

			}

		}

		Type ans = 0;

		for (int i = 0; i < n; i++) {

			//if (right[i] == -1) return -1;

			//if (g[i][right[i]] == -INF) return -1;

			ans += g[i][right[i]];

		}

		return -ans;

	}

} gao;

int t, n, m;

int main() {

	int cas = 0;

	scanf("%d", &t);

	while (t--) {

		scanf("%d%d", &n, &m);

		gao.init(n, n);

		int tmp;

		for (int i = 0; i < m; i++) {

			for (int j = 0; j < n; j++) {

				scanf("%d", &tmp);

				for (int k = 0; k < n; k++)

					gao.add_Edge(k, j, -abs(tmp - k - 1));

			}

		}

		printf("Case #%d: %d\n", ++cas, gao.km());

	}

	return 0;

}