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;
}