标签：find try final ota 赋值 contain int 思路 people
InputThe 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
OutputFor 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
思路：题意就是两个人选择一个接头地点，两个人去见面，然后要求两个人到达那个地点的路程之和最小。
可以分别计算两个人到达每个kfc的距离，用广度搜索计算，然后从两人都能到达的kfc中选取路程最小的那个。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 6 using namespace std; 7 8 typedef struct{ 9 int i,j; 10 int step; 11 }loca; 12 13 int n,m; 14 char mm[205][205];//地图 15 int flag[205][205];//此位置是否待过 16 int step[205][205];//到达此处的步数，只有@时才赋值 17 int isarrival[205][205];//如果此处两个人都到达才可以 18 int yifeni=0,yifenj=0;//yifen位置 19 int meri=0,merj=0;//mer位置 20 queue<loca> q; 21 loca now,nextt; 22 int add[2][4]={-1,1,0,0, 23 0,0,-1,1}; 24 25 void go(int si,int sj){ 26 now.i=si; 27 now.j=sj; 28 now.step=0; 29 flag[si][sj]=1; 30 q.push(now); 31 while(!q.empty()){ 32 now=q.front(); q.pop(); 33 if(mm[now.i][now.j]==‘@‘){ 34 step[now.i][now.j]+=now.step; 35 isarrival[now.i][now.j]++; 36 } 37 flag[now.i][now.j]=1; 38 for(int i=0;i<4;i++){ 39 nextt.i=now.i+add[0][i]; 40 nextt.j=now.j+add[1][i]; 41 nextt.step=now.step+1; 42 if(nextt.i<0||nextt.i>=n||nextt.j<0||nextt.j>m){ 43 continue; 44 } 45 if(flag[nextt.i][nextt.j]==0 && mm[nextt.i][nextt.j]!=‘#‘){ 46 q.push(nextt); 47 flag[nextt.i][nextt.j]=1; 48 } 49 } 50 } 51 } 52 53 int main() 54 { 55 while(~scanf("%d %d",&n,&m)){ 56 memset(step,0,sizeof(step)); 57 memset(isarrival,0,sizeof(isarrival)); 58 memset(flag,0,sizeof(flag)); 59 for(int i=0;i<n;i++){ 60 getchar(); 61 for(int j=0;j<m;j++){ 62 scanf("%c",&mm[i][j]); 63 if(mm[i][j]==‘Y‘){ 64 yifeni=i; yifenj=j; 65 } 66 if(mm[i][j]==‘M‘){ 67 meri=i; merj=j; 68 } 69 } 70 } 71 go(yifeni,yifenj); 72 memset(flag,0,sizeof(flag)); 73 go(meri,merj); 74 75 int minn=0x3f3f3f3f; 76 for(int i=0;i<n;i++){ 77 for(int j=0;j<m;j++){ 78 if(step[i][j]!=0&&isarrival[i][j]==2){ 79 int temp=step[i][j]; 80 if(temp<minn){ 81 minn=temp; 82 } 83 } 84 } 85 } 86 printf("%d\n",minn*11); 87 } 88 return 0; 89 }
标签：find try final ota 赋值 contain int 思路 people
原文地址：https://www.cnblogs.com/TWS-YIFEI/p/8821428.html
