HDU - 2757 Ocean Currents （BFS+优先队列）

来源：互联网发布：安卓必备软件编辑：程序博客网时间：2024/06/09 16:39

题意：求从开始点到目标点的最小的能量消耗，如果下一格的风向和当前的一样的话，消耗为0，否则为1，一共8个方向，其实方向是0

思路：BFS的基础上还要加上优先队列来取能量的最小值

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int MAXN = 1010;const int INF = 0x3f3f3f3f;struct node{int x,y,c;bool operator <(const node &a)const{return c > a.c;}}tmp;int n,m,sx,sy,ex,ey;char map[MAXN][MAXN];int f[MAXN][MAXN];int dir[8][2]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};void bfs(){priority_queue<node> q;node a = {sx,sy,0};q.push(a);while (!q.empty()){tmp = q.top();q.pop();int tx = tmp.x,ty = tmp.y;if (tx == ex && ty == ey)return;int cost;for (int i = 0; i < 8; i++){tx = tmp.x + dir[i][0];ty = tmp.y + dir[i][1];if (tx >= 0 && tx < n && ty >= 0 && ty < m){int d = (map[tmp.x][tmp.y]-'0');if (i == d)cost = tmp.c;else cost = tmp.c + 1;node cur = {tx,ty,cost};if (f[tx][ty] > cost){f[tx][ty] = cost;q.push(cur);}}}}}int main(){while (scanf("%d%d",&n,&m) != EOF){for (int i = 0; i < n; i++)scanf("%s",map[i]);int t;scanf("%d",&t);while (t--){for (int i = 0; i <= n; i++)for (int j = 0; j <= m; j++)f[i][j] = INF;scanf("%d%d%d%d",&sx,&sy,&ex,&ey);sx--,sy--,ex--,ey--;f[sx][sy] = 0;bfs();printf("%d\n",f[ex][ey]);}}return 0;}

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