Find a way
Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki. Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.
The input contains multiple test cases. Each test case include, first two integers n, m. (2<=n,m<=200). Next n lines, each line included m character. ‘Y’ express yifenfei initial position. ‘M’ express Merceki initial position. ‘#’ forbid road; ‘.’ Road. ‘@’ KCF
For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.
4 4 Y.#@ .... .#.. @..M 4 4 Y.#@ .... .#.. @#.M 5 5 Y..@. .#... .#... @..M. #...#
66 88 66
分别以两个人为起点进行两次BFS,然后找出步数最少的kfc店.
开始的时候我存了所有的kfc店位置,遍历kfc,然后BFS,这样就是BFS了(2*kfc店个数)次,超级无敌超时了...
#include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; int n, m; struct node{ int x,y; int step; }; struct node yifenfei; struct node friends; queue<struct node> q; int kfc[207][207]; int kind = 0;//kfc个数 char mp[207][207]; //地图 int vis[207][207]; int nnext[4][2]={{1,0}, {-1,0}, {0,1}, {0,-1}}; void bfs() { while(!q.empty())q.pop(); memset(vis,0,sizeof(vis)); node ynew; q.push(yifenfei); vis[yifenfei.x][yifenfei.y] = 1; while(q.size()) { ynew = q.front(); q.pop(); if(mp[ynew.x][ynew.y] == '@') { kfc[ynew.x][ynew.y] = ynew.step; } for(int i = 0; i < 4; i++) { int tx = ynew.x + nnext[i][0]; int ty = ynew.y + nnext[i][1]; if(tx >= 0 && tx < n && ty >= 0 && ty < m && mp[tx][ty] != '#' && vis[tx][ty] == 0) { vis[tx][ty] = 1; node f; f.x = tx; f.y = ty; f.step = ynew.step + 11; q.push(f); } } } while(!q.empty())q.pop(); memset(vis,0,sizeof(vis)); q.push(friends); vis[friends.x][friends.y] = 1; while(!q.empty()) { ynew = q.front(); q.pop(); if(mp[ynew.x][ynew.y] == '@') { kfc[ynew.x][ynew.y] += ynew.step; } for(int i = 0; i < 4; i++) { int tx = ynew.x + nnext[i][0]; int ty = ynew.y + nnext[i][1]; if(tx >= 0 && tx < n && ty >= 0 && ty < m && mp[tx][ty] != '#' && vis[tx][ty] == 0) { vis[tx][ty] = 1; node f; f.x = tx; f.y = ty; f.step = ynew.step + 11; q.push(f); } } } return; } int main() { while(scanf("%d%d", &n, &m) !=EOF && n && m) { memset(kfc, 0, sizeof(kfc)); for(int i = 0; i < n; i++) { getchar(); scanf("%s", mp[i]); for(int j = 0; j < m; j++) { if(mp[i][j] == 'Y') { yifenfei.x = i; yifenfei.y = j; yifenfei.step = 0; } else if(mp[i][j] == 'M') { friends.x = i; friends.y = j; friends.step = 0; } } } bfs(); int res = 1000000000; for(int i = 0; i < n; i++){ for(int j = 0; j < m; j++){ if(kfc[i][j] < res && kfc[i][j] != 0) res = kfc[i][j]; } } printf("%d\n", res); } return 0; }
