usaco 2.1

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进入第二章：上来就是一个比较繁琐的图论，求迷宫的最大联通块，然后允许拆一块墙，在求一次此时的最大联通块

求联通块：dfs+枚举每一块墙，要注意的是题目说要尽量靠西的答案，和尽量靠南的答案，那么在枚举墙的时候应

for (int j = 1; j < M; j++)

for (int i = N; i > 0; i--)

通过自底向上，从左到右枚举就省去了比较

2：每面墙会在两个小块内被表示：那么枚举时按题目要求,每个小块只用看他的N，E墙即可。

3：由于2，要将一面墙“打破”，将两个相邻块的bool数组置为false

/*ID: zhangw31PROG: castleLANG: C++*/#include <iostream>#include <fstream>#include <cstring>#include <algorithm>using namespace std;ifstream fin("castle.in");ofstream fout("castle.out");int M, N;int maxspace, number, tmp, maxs;const int MAXN = 55;bool vis[MAXN][MAXN];bool sta[MAXN][MAXN][4];int posx, posy;char direc;void check(int i, int j, int n){if (n % 2 == 1) {sta[i][j][2] = true; n -= 1;}if (n >= 8) {sta[i][j][1] = true;n -= 8;}if (n >= 4) {sta[i][j][3] = true;n -= 4;}if (n == 2) {sta[i][j][0] = true;}}void dfs(int i, int j){vis[i][j] = true;tmp ++;if (!sta[i][j][0] && !vis[i-1][j])  dfs(i-1, j);if (!sta[i][j][1] && !vis[i+1][j])  dfs(i+1, j);if (!sta[i][j][2] && !vis[i][j-1])  dfs(i, j-1);if (!sta[i][j][3] && !vis[i][j+1])  dfs(i, j+1);}void solve(){tmp = 0;for (int i = 1; i <= N; i++) {for (int j = 1; j <= M; j++) {if (!vis[i][j]) {number ++;dfs(i, j);if (tmp > maxspace) maxspace = tmp;tmp = 0;}}}}void reset(){for (int i = 1; i <= N; i++) {for (int j = 1; j <= M; j++) {vis[i][j] = false;}}for (int j = 0; j <= M + 1; j++) {vis[0][j] = true;vis[N+1][j] = true;}for (int i = 0; i <= N + 1; i++) {vis[i][0] = true;vis[i][M+1] = true;}}int main(){fin >> M >> N;for (int i = 1; i <= N; i++) {for (int j = 1; j <= M; j++) {fin >> tmp;check(i, j, tmp);}}solve();fout << number << endl;fout << maxspace << endl;for (int j = 1; j <= M; j++) {for (int i = N; i >= 1; i--) {if (sta[i][j][0]) {sta[i][j][0] = sta[i-1][j][1] = false;reset();solve();if (maxs < maxspace) {maxs = maxspace;posx = i; posy = j; direc = 'N';}sta[i][j][0] = sta[i-1][j][1] = true;}if (sta[i][j][3]) {sta[i][j][3] = sta[i][j+1][2] = false;reset();solve();if (maxs < maxspace) {maxs = maxspace;posx = i; posy = j; direc = 'E';}sta[i][j][3] = sta[i][j+1][2] = true;}}}fout << maxs << endl;fout << posx << " " << posy << " " << direc << endl;}

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