CodeForces 429B（）

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bfs的灵活运用

#include<iostream>#include<cstdio>#include<queue>#include<algorithm>using namespace std;const long N = 200000;struct v{int x, y;};int n, m, time;int step[4][2][2] = { { 0, 1, 1, 0 }, { 1, 0, 0, -1 }, { -1, 0, 0, 1 }, { -1, 0, 0, -1 } };int dis[4][1005][1005], visit[1005][1005], Map[1005][1005];bool ok(v a){if (a.x >= 1 && a.x <= n&&a.y >= 1 && a.y <= m){return true;}return false;}void bfs(int dir, int cx, int cy){queue<v>q;q.push(v{ cx, cy });visit[cx][cy] = time;dis[dir][cx][cy] = Map[cx][cy];while (!q.empty()){v ve = q.front();q.pop();for (int i = 0; i < 2; i++){v t = ve;t.x += step[dir][i][0]; t.y += step[dir][i][1];if (!ok(t))continue;dis[dir][t.x][t.y] = max(dis[dir][ve.x][ve.y] + Map[t.x][t.y], dis[dir][t.x][t.y]);if (visit[t.x][t.y] == time)continue;q.push(t);visit[t.x][t.y] = time;}}}int main(){cin >> n >> m;for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)scanf("%d", &Map[i][j]);time = 0;time++; bfs(0, 1, 1);time++; bfs(1, 1, m);time++; bfs(2, n, 1);time++; bfs(3, n, m);int mm = 0;for (int i = 2; i <= n - 1; i++)for (int j = 2; j <= m - 1; j++)mm = max(mm, max(dis[0][i - 1][j] + dis[1][i][j + 1] + dis[2][i][j - 1] + dis[3][i + 1][j],dis[1][i - 1][j] + dis[0][i][j - 1] + dis[3][i][j + 1] + dis[2][i + 1][j]));cout << mm << endl;return 0;}

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