Hdu 2757 Ocean Currents

来源：互联网发布：qq农场刷经验软件编辑：程序博客网时间：2024/06/06 02:58

大意不再赘述。

思路：比较简单的广搜，给定8个方向，只要现在广搜的方向与之前所在的点的风的方向不同的话，消耗量加1即可。可以用优先队列每次取出时间最少的那个点，用TIME数组判状态，如简单的判是否走过的话，会TLE，而用TIME数组保证最先到达终点的点一定是消耗量最小的。

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <string>#include <queue>using namespace std;const int MAXN = 1010;const int INF = 0x3f3f3f3f;const int dx[] = {-1, -1, 0, 1, 1, 1, 0, -1};const int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};int n, m;struct node{int x, y;int step;bool operator < (const node &a) const{return a.step < step;}};char maze[MAXN][MAXN];int TIME[MAXN][MAXN];int check(int x, int y){if(x >= 1 && x <= n && y >= 1 && y <= m) return 1;return 0;}priority_queue<node> Q;void init(){for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){TIME[i][j] = INF;}}}void bfs(int bx, int by, int ex, int ey){while(!Q.empty()) Q.pop();node cur, next;cur.x = bx, cur.y = by, cur.step = 0;TIME[bx][by] = 0;Q.push(cur);while(!Q.empty()){cur = Q.top(); Q.pop();if(cur.x == ex && cur.y == ey) return ;for(int i = 0; i < 8; i++){next = cur;next.x += dx[i], next.y += dy[i];if(i != maze[cur.x][cur.y]-'0') next.step++; //方向与以前不同，就消耗量加1 if(check(next.x, next.y) && next.step < TIME[next.x][next.y]){TIME[next.x][next.y] = next.step;Q.push(next);}}}return ;}void read_case(){for(int i = 1; i <= n; i++) scanf("%s", maze[i]+1);}void solve(){read_case();int q;scanf("%d", &q);while(q--){init();int bx, by, ex, ey;scanf("%d%d%d%d", &bx, &by, &ex, &ey);bfs(bx, by, ex, ey);printf("%d\n", TIME[ex][ey]);}return ;}int main(){while(~scanf("%d%d", &n, &m)){solve();}return 0;}