UVA 11383 - Golden Tiger Claw

題目鏈接

題意：給定每列和每行的和，給定一個矩陣，要求每一個格子(x, y)的值小于row(i) + col(j)，求一種方案，而且全部行列之和的和最小

思路：A二分圖完美匹配的擴展，行列建二分圖，權值為矩陣對應位置的值，做一次KM算法后。全部頂標之和就是最小的

代碼：

#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; }

轉載于:https://www.cnblogs.com/gccbuaa/p/6896235.html

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