uva 10615 - Rooks（完美匹配）

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;
}