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; }