Simple HDU 2612 Find a way,bfs.

Description

Pass a year learning in Hangzhou, Yifenfei arrival hometown Ningbo at finally. Leave Ningbo One year, Yifenfei has many people to meet. Especially a good friend Merceki.
Yifenfei's home is on the countryside, but Merceki's home is in the center of the city. So Yifenfei made arrangements and Merceki to meet at a KFC. There is many KFC in Ningbo, they want to choose one and the total time to it is most smallest.
Now give your a Ningbo map, Both Yifenfei and Merceki can move up, down, left, right to the adjacent road by cost one minute S.

is to ask two people to the minimum value of a KFC, this question remember to do when the pit was miserable, to pay attention to the initialization of the INF, and said that the starting position of people to be expressed as a pass.

The code is as follows:

#include <iostream>#include<cstring>using namespacestd;Const intinf=1e7; Shortmap1[205][205];intque[40005],las,fir;intans[205][205];intans1[205][205];intn,m;intCoukfc,si,sj,ei,ej;BOOLJudgeintXintYint(*rem) [205]){    if(x<=0|| y<=0|| x>n| | Y>M)return 0; if(rem[x][y]!=INF)return 0; if(map1[x][y]==0)        return 0; return 1;}voidBFsintXintYint(*rem) [205]) {las=fir=0; intcou=0; inttemp,t1,t2; Que[las++]=x* ++y; Rem[x][y]=0;  while(las-fir) {Temp=que[fir++]; T1=temp/ +; T2=temp% +; Temp=REM[T1][T2]; if(map1[t1][t2]==2)            ++cou; if(cou>=COUKFC)return; --T1; if(judge (T1,t2,rem)) {Rem[t1][t2]=temp+1; Que[las++]=t1* ++T2; } T1+=2; if(judge (T1,t2,rem)) {Rem[t1][t2]=temp+1; Que[las++]=t1* ++T2; }            --T1; --T2; if(judge (T1,t2,rem)) {Rem[t1][t2]=temp+1; Que[las++]=t1* ++T2; } T2+=2; if(judge (T1,t2,rem)) {Rem[t1][t2]=temp+1; Que[las++]=t1* ++T2; }    }}intSlove () {BFS (Si,sj,ans);    BFS (EI,EJ,ANS1); intminn=INF;  for(intI=1; i<=n;++i) for(intj=1; j<=m;++j)if(map1[i][j]==2)                if(minn>ans[i][j]+ANS1[I][J]) Minn=ans[i][j]+Ans1[i][j]; returnminn* One;}intMain () {Ios::sync_with_stdio (false); CharC;  while(cin>>n>>M) {COUKFC=0;  for(intI=1; i<=n;++i) for(intj=1; j<=m;++j) {cin>>C; ANS[I][J]=ans1[i][j]=INF; Switch(c) { Case 'Y': Map1[i][j]=1; Si=i; Sj=J;  Break;  Case 'M': Map1[i][j]=1; Ei=i; Ej=K;  Break;  Case '.': Map1[i][j]=1;  Break;  Case '#': Map1[i][j]=0;  Break;  Case '@': Map1[i][j]=2; ++COUKFC;  Break; }} cout<<slove () <<Endl; }    return 0;}

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Simple HDU 2612 Find a way,bfs.