uva 10615 - Rooks（完美匹配）

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题目链接：uva 10615 - Rooks

显而易见，需要用到颜色种类即为行列中棋子个数的最大值k。问题是如何构造，建图，行和列去匹配，每次匹配一种颜色，将匹配到的边删除即可。需要注意的是，每次需要删除k条边，所以对于度数不足的点需要用无效边凑数，这样保证最后答案可以构造出来。

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 105;int N, M, L[maxn], ans[maxn][maxn], row[maxn], col[maxn], in[maxn];bool S[maxn], T[maxn];vector<int> g[maxn];char G[maxn][maxn];bool match (int u) {for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (!T[v]) {T[v] = true;if (!L[v] || match(L[v])) {L[v] = u;return true;}}}return false;}void KM () {memset(L, 0, sizeof(L));for (int i = 1; i <= N; i++) {memset(T, false, sizeof(T));match(i);}}void init () {scanf("%d", &N);for (int i = 1; i <= N; i++)scanf("%s", G[i]+1);memset(ans, 0, sizeof(ans));memset(row, 0, sizeof(row));memset(col, 0, sizeof(col));for (int i = 1; i <= N; i++) {for (int j = 1; j <= N; j++) if (G[i][j] == '*') {row[i]++; col[j]++;}}M = 0;for (int i = 1; i <= N; i++)M = max(M, max(row[i], col[i]));}void solve () {memset(in, 0, sizeof(in));for (int i = 1; i <= N; i++) {g[i].clear();for (int j = 1; j <= N; j++) if (G[i][j] == '*') {g[i].push_back(j);in[j]++;}}for (int i = 1; i <= N; i++) {for (int j = 1; j <= N && g[i].size() < M; j++) {while (in[j] < M && g[i].size() < M) {g[i].push_back(j);in[j]++;}}}for (int c = 1; c <= M; c++) {KM();for (int i = 1; i <= N; i++) {int u = L[i];if (G[u][i] == '*') ans[u][i] = c;for (int j = 0; g[u].size(); j++) if (g[u][j] == i) {g[u].erase(g[u].begin() + j);break;}}}printf("%d\n", M);for (int i = 1; i <= N; i++) {for (int j = 1; j <= N; j++)printf("%d%c", ans[i][j], j == N ? '\n' : ' ');}}int main () {int cas;scanf("%d", &cas);while (cas--) {init();solve();}return 0;}

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