題目鏈接：http://acm.hdu.edu.cn/showPRoblem.php?pid=2612 題意：y和m要在kfc見面，問你能否選擇一家kfc使得y和m到那里的距離最短 解析：兩個人都跑一遍bfs，然后枚舉每家kfc，找出最優的即可
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>using namespace std;const int inf = 0x7fffffff;int dx[] = {0,1,-1,0};int dy[] = {1,0,0,-1};int vis[205][205];int step1[205][205];int step2[205][205];int n,m;char a[205][205];struct node{ int x,y; int step; node() {} node(int _x,int _y,int _step) { x = _x; y = _y; step = _step; }};void bfs(int sx,int sy,int step[205][205]){ memset(vis,0,sizeof(vis)); queue<node>q; q.push(node(sx,sy,0)); while(!q.empty()) { node now = q.front(); q.pop(); if(a[now.x][now.y]=='@') { if(step[now.x][now.y]) step[now.x][now.y] = min(step[now.x][now.y],now.step); else step[now.x][now.y] = now.step; } for(int i=0;i<4;i++) { int tx = now.x+dx[i]; int ty = now.y+dy[i]; if(tx<0 || tx>=n || ty<0 || ty>=m) continue; if(a[tx][ty]=='#' || vis[tx][ty]) continue; vis[tx][ty] = 1; q.push(node(tx,ty,now.step+1)); } }}int main(){ while(~scanf("%d %d",&n,&m)) { int x1,y1,x2,y2; for(int i=0;i<n;i++) { scanf("%s",a[i]); for(int j=0;j<m;j++) { if(a[i][j]=='Y') { x1 = i; y1 = j; } if(a[i][j]=='M') { x2 = i; y2 = j; } } } memset(step1,0,sizeof(step1)); memset(step2,0,sizeof(step2)); bfs(x1,y1,step1); bfs(x2,y2,step2); int ans = inf; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { if(a[i][j] == '@') { if(step1[i][j] && step2[i][j]) ans = min(ans,step1[i][j]+step2[i][j]); } } } printf("%d/n",ans*11); } return 0;}新聞熱點
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