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UVA 11383 - Golden Tiger Claw(二分图完美匹配扩展)
UVA 11383 - Golden Tiger Claw
题目链接
题意:给定每列和每行的和,给定一个矩阵,要求每个格子(x, y)的值小于row(i) + col(j),求一种方案,并且所有行列之和的和最小
思路:A二分图完美匹配的扩展,行列建二分图,权值为矩阵相应位置的值,做一次KM算法后,所有顶标之和就是最小的
代码:
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int MAXNODE = 505; typedef int Type; const Type INF = 0x3f3f3f3f; struct KM { int n; Type g[MAXNODE][MAXNODE]; Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE]; int left[MAXNODE]; bool S[MAXNODE], T[MAXNODE]; void init(int n) { this->n = n; } 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 < n; 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; return true; } } else slack[j] = min(slack[j], tmp); } return false; } void update() { Type a = INF; for (int i = 0; i < n; i++) if (!T[i]) a = min(a, slack[i]); for (int i = 0; i < n; i++) { if (S[i]) Lx[i] -= a; if (T[i]) Ly[i] += a; } } void km() { for (int i = 0; i < n; i++) { left[i] = -1; Lx[i] = -INF; Ly[i] = 0; for (int j = 0; j < n; j++) Lx[i] = max(Lx[i], g[i][j]); } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) slack[j] = INF; while (1) { for (int j = 0; j < n; j++) S[j] = T[j] = false; if (dfs(i)) break; else update(); } } } } gao; int n; int main() { while (~scanf("%d", &n)) { gao.init(n); for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) { scanf("%d", &gao.g[i][j]); } gao.km(); int ans = 0; for (int i = 0; i < n; i++) { printf("%d%c", gao.Lx[i], i == n - 1 ? '\n' : ' '); ans += gao.Lx[i]; } for (int i = 0; i < n; i++) { printf("%d%c", gao.Ly[i], i == n - 1 ? '\n' : ' '); ans += gao.Ly[i]; } printf("%d\n", ans); } return 0; }
UVA 11383 - Golden Tiger Claw(二分图完美匹配扩展)
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