hdu 5556 Land of Farms（二分图匹配）

解题思路

代码

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>

using namespace std;
typedef pair<int,int> pii;
const int maxn = 105;
const int dir[][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

int N, M, K, ans, idx[maxn][maxn];
bool used[maxn], xset[maxn];
char str[maxn][maxn];
vector<int> G[maxn];
vector<pii> T[20];

void init () {
K = 0;
memset(idx, 0, sizeof(idx));
memset(xset, false, sizeof(xset));
for (int i = 0; i < 10; i++) T[i].clear();

scanf("%d%d", &N, &M);
for (int i = 0; i < N; i++) {
scanf("%s", str[i]);
for (int j = 0; j < M; j++) {
if (str[i][j] == '.') {
idx[i][j] = ++K;
if ((i+j) % 2 == 1)
xset[K] = true;
} else
T[str[i][j] - '0'].push_back(make_pair(i, j));
}
}

for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (str[i][j] == '.' && ((i+j)&1)) {
int u = idx[i][j];
G[u].clear();
for (int d = 0; d < 4; d++) {
int x = i + dir[d][0];
int y = j + dir[d][1];
if (x < 0 || x >= N || y < 0 || y >= M || idx[x][y] == 0) continue;
G[u].push_back(idx[x][y]);
}
}
}
}
}

void draw(int u) {
for (int i = 0; i < T[u].size(); i++) {
int x = T[u][i].first, y = T[u][i].second;
for (int d = 0; d < 4; d++) {
int p = x + dir[d][0], q = y + dir[d][1];
if (p < 0 || p >= N || q < 0 || q >= M || idx[p][q] == 0) continue;
used[idx[p][q]] = true;
}
}
}

int distance(int a, int b) {
int ret = 0x3f3f3f3f;
for (int i = 0; i < T[a].size(); i++) {
for (int j = 0; j < T[b].size(); j++)
ret = min(ret, abs(T[a][i].first-T[b][j].first) + abs(T[a][i].second-T[b][j].second));
}
return ret;
}

bool judge (int s, int u) {
for (int i = 0; i < u; i++) {
if ((s&(1<<i)) && distance(i, u) <= 1)
return false;
}
return true;
}

int L[maxn];
bool t[maxn];

bool match(int u) {
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (used[v]) continue;
if (!t[v]) {
t[v] = true;
if (!L[v] || match(L[v])) {
L[v] = u;
return true;
}
}
}
return false;
}

int KM () {
int ret = 0;
memset(L, 0, sizeof(L));
for (int i = 1; i <= K; i++) {
if (used[i] || !xset[i]) continue;
memset(t, false, sizeof(t));
if (match(i)) ret++;
}
return ret;
}

int solve (int ti) {
int n = 0;
for (int i = 1; i <= K; i++) if (used[i]) n++;
return K - n - KM();
}

void dfs (int d, int s, int add) {

if (d >= 10) {
ans = max(ans, solve(add) + add);
return;
}

bool tmp[maxn];
memcpy(tmp, used, sizeof(used));

if (T[d].size() && judge(s, d)) {
draw(d);
dfs(d + 1, s|(1<<d), add + 1);
memcpy(used, tmp, sizeof(tmp));
}

dfs(d + 1, s, add);
memcpy(used, tmp, sizeof(tmp));
}

int main () {
int cas;
scanf("%d", &cas);
for (int kcas = 1; kcas <= cas; kcas++) {
init();

ans = 0;
memset(used, 0, sizeof(used));
dfs(0, 0, 0);

printf("Case #%d: %d\n", kcas, ans);
}
return 0;
}