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Find a way

Time Limit: 3000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21857????Accepted Submission(s): 7120


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.

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

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 4Y.#@.....#..@..M4 4Y.#@.....#..@#.M5 5Y..@..#....#...@..M.#...# Sample Output 668866 Author yifenfei
Source 奮斗的年代
Recommend yifenfei???|???We have carefully selected several similar problems for you:??2717?1254?1728?2102?1072

題意：

給一張地圖，Y代表第一個(gè)人所在的位置，M代表第二個(gè)人所在的位置，.代表路，#代表墻，@代表KFC， 兩個(gè)人想在同一個(gè)kfc見(jiàn)面，問(wèn)他們見(jiàn)面花費(fèi)最小的時(shí)間；

分析：

遍歷地圖，對(duì)于每一KFC，分別求出Y到這個(gè)KFC的距離dy，M到這個(gè)KFC的距離dm， 如果距離為dy=-1或者dm=-1，則說(shuō)明無(wú)法到達(dá)； 否則，跟新最短距離：ans=min(ans,dy+dm);

這個(gè)思路比較簡(jiǎn)單，但是當(dāng)KFC的數(shù)量非常多的時(shí)候，就會(huì)超時(shí)

根據(jù)上面的思路，我們可以優(yōu)化一下，減少重復(fù)計(jì)算，設(shè)置一個(gè)距離數(shù)組dis[N][N][2];初始化為INF

dis[i][j][0]代表Y到坐標(biāo)為（i，j）的KFC的距離，dis[i][j][1]代表M到坐標(biāo)為（i，j）的KFC的距離

然后，BFS求出Y到地圖上所有點(diǎn)的最短距離，若這個(gè)點(diǎn)是KFC，則更新dis[i][j][0],

同理，BFS求出M到地圖上所有點(diǎn)的最短距離，若這個(gè)點(diǎn)是KFC，則更新dis[i][j][1],

最后遍歷地圖，如果這個(gè)點(diǎn)是KFC，且當(dāng)前最短距離 > Y到這個(gè)KFC的距離+M到這個(gè)KFC的距離，

則更新最短距離，最后不要忘記*11哦

超時(shí)代碼

#include<iostream> #include<cstring> #include<queue> #include<cmath> #include<cstdio> using namespace std; #define MAXV 209 char mp[MAXV][MAXV];//地圖 bool vis[MAXV][MAXV];//判斷點(diǎn)是否在隊(duì)列 int n,m; struct point {int x,y,step; }; point Y,M;//兩人的位置 vector<point> KFC;//儲(chǔ)存KFC的位置 int dic[4][2]={0,1,0,-1,1,0,-1,0}; bool check(int x,int y)//若這一點(diǎn)可以走，返回0 {if(x<0||x>=n||y<0||y>=m||vis[x][y]||mp[x][y]=='#')return 1;return 0; } int bfs(point s,point e)//求s點(diǎn)到e點(diǎn)的最短距離，無(wú)法到達(dá)距離為-1 {memset(vis,0,sizeof(vis));queue<point>q;q.push(s);//起點(diǎn)入隊(duì)vis[s.x][s.y]=1;while(!q.empty()){point now=q.front();q.pop();if(now.x==e.x&&now.y==e.y)//已經(jīng)到達(dá)終點(diǎn)return now.step;for(int i=0;i<4;i++)//上下左右，四個(gè)方向搜索{int x=now.x+dic[i][0];int y=now.y+dic[i][1];if(check(x,y))//不合法continue;point next;next.x=x;next.y=y;next.step=now.step+1;vis[x][y]=1;q.push(next);}}return -1;//無(wú)法到達(dá) } int main() {while(scanf("%d%d",&n,&m)!=EOF){KFC.clear();for(int i=0;i<n;i++){scanf("%s",mp[i]);for(int j=0;j<m;j++){if(mp[i][j]=='Y'){Y.x=i;Y.y=j;Y.step=0;}if(mp[i][j]=='M'){M.x=i;M.y=j;M.step=0;}if(mp[i][j]=='@'){point temp;temp.x=i;temp.y=j;temp.step=0;KFC.push_back(temp);}}}int ans=1<<20;for(int i=0;i<KFC.size();i++)//遍歷所有KFC{int dy=bfs(Y,KFC[i]),dm=bfs(M,KFC[i]);if(dy==-1||dm==-1)continue;ans=min(ans,dy+dm);}cout<<ans*11<<endl;}return 0; }
AC代碼
#include<iostream> #include<queue> #include<cstdio> #include<cstring> using namespace std; const int N = 210; const int inf = 100000000; int n, m, flag;//flag=0代表Y，flag=1代表M int dis[N][N][2]; //dis[i][j][0]代表Y到坐標(biāo)為（i，j）的KFC的距離 //dis[i][j][1]代表M到坐標(biāo)為（i，j）的KFC的距離 int mark[N][N];//標(biāo)記點(diǎn)是否在隊(duì)列中 int dir[4][2] = {{1, 0}, {0, -1}, {0, 1}, {-1, 0}};//搜索方向 char s[N][N];//地圖 struct node {int x, y, step; }; void bfs(int x, int y) {queue<node>q;node temp, type;temp.x = x;temp.y = y;temp.step = 0;q.push(temp);//起點(diǎn)入隊(duì)mark[x][y] = 1;//標(biāo)記while(!q.empty()){temp = q.front();q.pop();//出隊(duì)type.step = temp.step + 1;//步數(shù)加一for(int i = 0; i < 4; i++){type.x = x = temp.x + dir[i][0];type.y = y = temp.y + dir[i][1];if(x >= 0 && x < n && y >= 0 && y < m && mark[x][y] == 0 && s[x][y]!='#')//可以走{mark[x][y] = 1;if(s[x][y] == '@')dis[x][y][flag] = type.step;//距離q.push(type);}}} } int main() {while(scanf("%d%d", &n, &m)!=EOF){int min = inf;for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)dis[i][j][0] = dis[i][j][1] = inf;//初始化距離為INFfor(int i = 0; i < n; i++)scanf("%s", s[i]);for(int i = 0; i < n; i++)for(int j = 0; j < m; j++){if(s[i][j] == 'Y'){flag = 0;memset(mark, 0, sizeof(mark));bfs(i, j);}if(s[i][j] == 'M'){flag = 1;memset(mark, 0, sizeof(mark));bfs(i, j);}}for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)if(s[i][j] == '@' && min > dis[i][j][0] + dis[i][j][1])//松馳操作{min = dis[i][j][0] + dis[i][j][1];//更新最短距離}printf("%d\n", min*11);} }

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