HDU 2853 Assignment

题目链接

题意：如今有N个部队和M个任务（M>=N），每一个部队完毕每一个任务有一点的效率，效率越高越好。可是部队已经安排了一定的计划，这时须要我们尽量用最小的变动，使得全部部队效率之和最大。求最小变动的数目和变动后和变动前效率之差。

思路：对于怎样保证改变最小，没思路，看了别人题解，恍然大悟，表示想法很机智

试想，假设能让原来那些匹配边，比其它匹配出来总和同样的权值还大，对结果又不影响，那就简单了，这个看似不能做到，事实上是能够做到的

数字最多选出50个，所以把每一个数字乘上一个大于50的数字k，然后原来匹配的权值多+1，这样每k倍数字代表了原来的1，而求出来的即使有原来的多匹配的，总权值等于也不会多1，跟原来的结果一样，又保证了跟其它匹配方式相比，这个优先级更大，很的巧妙

代码：

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

const int MAXNODE = 55;

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;

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

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

				g[i][j] = -INF;

	}

	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;

	}

	int to[MAXNODE];

	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] == to[i]) ans += (g[i][right[i]] - 1) / (n + 1);

			else ans += g[i][right[i]] / (n + 1);

		}

		return ans;

	}

	void solve() {

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

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

				scanf("%d", &g[i][j]);

				g[i][j] *= (n + 1);

			}

		int pre = 0;

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

			scanf("%d", &to[i]);

			to[i]--;

			pre += g[i][to[i]] / (n + 1);

			g[i][to[i]]++;

		}

		int cnt = 0;

		int ans = km() - pre;

		for (int i = 0; i < n; i++) if (right[i] != to[i]) cnt++;

		printf("%d %d\n", cnt, ans);

	}

} gao;

int n, m;

int main() {

	while (~scanf("%d%d", &n, &m)) {

		gao.init(n, m);

		gao.solve();

	}

	return 0;

}