HDU 2853 Assignment
题意:现在有N个部队和M个任务(M>=N),每个部队完成每个任务有一点的效率,效率越高越好。但是部队已经安排了一定的计划,这时需要我们尽量用最小的变动,使得所有部队效率之和最大。求最小变动的数目和变动后和变动前效率之差。
思路:对于如何保证改变最小,没思路,看了别人题解,恍然大悟,表示想法非常机智
试想,如果能让原来那些匹配边,比其他匹配出来总和相同的权值还大,对结果又不影响,那就简单了,这个看似不能做到,其实是可以做到的
数字最多选出50个,所以把每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,又保证了跟其他匹配方式相比,这个优先级更大,非常的巧妙
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; }
时间: 2024-10-08 20:50:20