http://acm.hdu.edu.cn/showproblem.php?pid=2612
Find a way
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32208 Accepted Submission(s): 10316
Problem Description
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.
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.
Input
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
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
Output
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.
Sample Input
4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#
Sample Output
66
88
66
Author
yifenfei
Source
Recommend
题意：分别从F,M出发去往某个@的最小距离。。
new note：这个码有bug但还是过了。。。没初始化ans数组inf
3 3 Y#@ .M# ..@
output： 66
//#include <bits/stdc++.h> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <algorithm> #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdio.h> #include <queue> #include <stack>; #include <map> #include <set> #include <string.h> #include <vector> #define ME(x , y) memset(x , y , sizeof(x)) #define SF(n) scanf("%d" , &n) #define rep(i , n) for(int i = 0 ; i < n ; i ++) #define INF 0x3f3f3f3f #define mod 1000000007 using namespace std; typedef long long ll ; char a[209][209]; int n , m ; int index[2] , indey[2]; int mx , my ; int vis[209][209]; int dis[4][2] = {{1 , 0} , {-1 , 0} , {0 , -1} , {0 , 1}}; int step1[2][209][209]; struct node { int x , y , step; node(int x , int y , int step):x(x),y(y),step(step){} node(){}; }; bool check(int x , int y) { if(x <= 0 || y <= 0 || x > n || y > m) return false ; if(!vis[x][y] && a[x][y] != '#') return true ; return false ; } int bfs(int x , int y , int p) { queue<node>q; node now , next ; memset(vis , 0 , sizeof(vis)); q.push(node(x , y , 0)); vis[x][y] = 1 ; step1[p][x][y] = 0 ; while(!q.empty()) { now = q.front(); q.pop(); for(int i = 0 ; i < 4 ; i++) { next.x = now.x + dis[i][0]; next.y = now.y + dis[i][1]; next.step = now.step + 1 ; if(check(next.x , next.y)) { vis[next.x][next.y] = 1 ; step1[p][next.x][next.y] = next.step ; q.push(next); } } } return 0 ; } int main() { while(~scanf("%d%d" , &n , &m)) { getchar(); for(int i = 1 ; i <= n ; i++) { for(int j = 1 ; j <= m ; j++) { scanf("%c" , &a[i][j]); if(a[i][j] == 'Y') { index[0] = i; indey[0] = j; } else if(a[i][j] == 'M') { index[1] = i; indey[1] = j; } } getchar(); } int ans = INF; bfs(index[0] , indey[0] , 0); bfs(index[1] , indey[1] , 1); for(int i = 1 ; i <= n ; i++) { for(int j = 1 ; j <= m ; j++) { if(a[i][j] == '@') { ans = min(ans , step1[0][i][j] + step1[1][i][j]); } } } printf("%d " , ans*11); } return 0; }
比赛后码：
//#include <bits/stdc++.h> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <cstdio> #include <string> #include <stdio.h> #include <queue> #include <stack> #include <map> #include <set> #include <string.h> #include <vector> #define ME(x , y) memset(x , y , sizeof(x)) #define SF(n) scanf("%d" , &n) #define rep(i , n) for(int i = 0 ; i < n ; i ++) #define INF 0x3f3f3f3f #define mod 1000000007 #define PI acos(-1) using namespace std; typedef long long ll ; const int N = 1e7 + 5 ; char s[209][209]; int vis[209][209]; int dir[4][2] = {{1 , 0} , {-1 , 0} , {0 , 1} , {0 , -1}}; int n , m ; int ans[2][209][209]; struct node{ int x , y , w; }; void bfs(int t , int x , int y) { node now , next , last; queue<node>q; now.x = x , now.y = y , now.w = 0 ; q.push(now); vis[x][y] = 1 ; while(!q.empty()) { next = q.front(); q.pop(); if(s[next.x][next.y] == '@') { ans[t][next.x][next.y] = next.w ; } for(int i = 0 ; i < 4 ; i++) { int xx = next.x + dir[i][0]; int yy = next.y + dir[i][1]; int nw = next.w + 1 ; if(xx < 0 || xx >= n || yy < 0 || yy >= m) { continue ; } if(vis[xx][yy] || s[xx][yy] == '#') { continue ; } vis[xx][yy] = 1 ; last.x = xx , last.y = yy , last.w = nw ; q.push(last); } } } int main() { while(~scanf("%d%d" , &n , &m)) { int x , y , x1 , y1 ; memset(vis , 0 , sizeof(vis)); memset(ans , INF , sizeof(ans));//没有考虑到达不了的@（但最少有一个@可达） //到达不了的@应该为无穷大，否则为0则答案错误 for(int i = 0 ; i < n ; i++) { for(int j = 0 ; j < m ; j++) { cin >> s[i][j] ; if(s[i][j] == 'Y') { x = i , y = j ; } if(s[i][j] == 'M') { x1 = i , y1 = j ; } } } bfs(0 , x , y); memset(vis , 0 , sizeof(vis)); bfs(1 , x1 , y1); int mi = INF ; for(int i = 0 ; i < n ; i++) { for(int j = 0 ; j < m ; j++) { if(s[i][j] == '@') { mi = min(mi , ans[0][i][j] + ans[1][i][j]); } } } cout << mi * 11 << endl ; } return 0 ; }