题目地址:点击打开链接
题意:小明要从一个矩阵的(0,0)点到(n-1,m-1)点问最少花费多少时间到达,
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.
思路:参考大神A的,学到几个技巧,刚开始把图转化为int类型,在后面不用每次判断,还有一点是这个和普通的BFS搜索题有点不同,一个人可以在一个点呆好几秒,这样必须就得用到优先队列,不然按出栈顺序到达的有可能不是最短的时间,还有就是判断条件一定要写清,因为这个wrong了几发,还有就是程序不能用next标记变量或数组,会导致编译错误,注意一下
AC代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> typedef long long ll; using namespace std; int n,m,temp; int map1[110][110]; int visit[110][110]; int pre[110][110]; char a[110]; struct node { int x,y; int step; friend bool operator < (node a,node b) { return a.step > b.step;//最小值优先 } }now,future; int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}}; bool judge(int x,int y) { if(x >= 0 && x < n && y >= 0 && y < m && !visit[x][y] && map1[x][y] != -1)//<n和<m加了等号,wrong了几发,太粗心了 return true; return false; } int bfs() { int i; priority_queue<node> pq; now.x = 0; now.y = 0; now.step = 0; visit[0][0] = 1; pq.push(now); while(!pq.empty()) { future = pq.top(); pq.pop(); if(future.x == n-1 && future.y == m-1) return future.step; for(i=0; i<4; i++) { now.x = future.x + dir[i][0]; now.y = future.y + dir[i][1]; if(judge(now.x,now.y)) { visit[now.x][now.y] = 1; now.step = future.step + 1 + map1[now.x][now.y]; pre[now.x][now.y] = i; pq.push(now); } } } return -1; } void shuchu(int x,int y) { if(pre[x][y] == -1) return; int newx = x - dir[pre[x][y]][0]; int newy = y - dir[pre[x][y]][1]; shuchu(newx,newy); printf("%ds:(%d,%d)->(%d,%d)\n",temp++,newx,newy,x,y); while(map1[x][y] > 0) { map1[x][y]--; printf("%ds:FIGHT AT (%d,%d)\n",temp++,x,y); } } int main() { int i,j; while(scanf("%d%d",&n,&m) != EOF) { memset(pre,-1,sizeof(pre)); memset(visit,0,sizeof(visit)); getchar(); for(i=0; i<n; i++) { gets(a); for(j=0; j<m; j++) { if(a[j] == '.') { map1[i][j] = 0; } else if(a[j] == 'X') { map1[i][j] = -1; } else if(a[j] >= '0' && a[j] <= '9') { map1[i][j] = a[j] - '0'; } } } int x = bfs(); if(x == -1) { printf("God please help our poor hero.\n"); } else { printf("It takes %d seconds to reach the target position, let me show you the way.\n",x); temp = 1; shuchu(n-1,m-1); } printf("FINISH\n"); } return 0; }