Bzoj 1698 [Usaco2007 feb]lilypad Pond Lotus leaf Pond bfs+ Shortest

Test Instructions:Link * * Method: **bfs+ Shortest circuit parsing:This problem is still very interesting. The first idea gets stuck, the first one is two cents +bfs, but if so, what can be done? To a point of traffic. So you have to change your mind. You may want to redefine the shortest path. Consider all 0 of the points in the diagram as feasible, and each time you search from that point to 0 points, and then treat this edge as 1. That is, f[i][j] represents the shortest circuit to i,j point. Clearly the end of the sentence, and then at the same time to record the shortest possible number. Code:

#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N#define INF 0x3f3f3f3fusing namespace STD;typedef Long LongllintN,m;int MapN [N];ll Num[n][n];intF[n][n];intV[n][n];structpoint{intx, y;} st,ed;BOOLCanarrive[n][n][n][n];intxx[9]={0,-1,-2,-2,-1,1,2,2,1};intyy[9]={0,-2,-1,1,2,2,1,-1,-2};voidFloodFill (intXintY) {memset(V,0,sizeof(v)); queue<point>Q    Point TMP;    Tmp.x=x,tmp.y=y; Q.push (TMP); while(!q.empty ())        {point U=q.front (); Q.pop (); for(intI=1; i<=8; i++) {tmp.x=u.x+xx[i],tmp.y=u.y+yy[i];if(tmp.x<=0|| tmp.y<=0|| tmp.x>n| | tmp.y>m| | V[TMP.X][TMP.Y])Continue;if(!Map[Tmp.x] [tmp.y]| |            (TMP.X==ED.X&AMP;&AMP;TMP.Y==ED.Y)) {canarrive[x][y][tmp.x][tmp.y]=1; }Else if(Map[Tmp.x] [tmp.y]==1) {Q.push (TMP); v[tmp.x][tmp.y]=1; }        }    }}voidBFS () {memset(V,0,sizeof(v)); queue<point>Q    Q.push (ST); v[st.x][st.y]=1; f[st.x][st.y]=0; num[st.x][st.y]=1; while(!q.empty ())        {point U=q.front ();        Q.pop (); v[u.x][u.y]=0; for(intI=1; i<=n;i++) { for(intj=1; j<=m;j++) {if(Canarrive[u.x][u.y][i][j]) {if(I==ED.X&AMP;&AMP;J==ED.Y) {if(F[u.x][u.y]<f[i][j])                            {F[I][J]=F[U.X][U.Y];                        NUM[I][J]=NUM[U.X][U.Y]; }Else if(F[u.x][u.y]==f[i][j])                        {NUM[I][J]+=NUM[U.X][U.Y]; }                    }Else{if(f[u.x][u.y]+1<f[i][j]) {f[i][j]=f[u.x][u.y]+1; NUM[I][J]=NUM[U.X][U.Y];if(!v[i][j]) {v[i][j]=1;                                Point TMP;                                Tmp.x=i,tmp.y=j;                            Q.push (TMP); }                        }Else if(f[u.x][u.y]+1==f[i][j]) {NUM[I][J]+=NUM[U.X][U.Y];if(!v[i][j]) {v[i][j]=1;                                Point TMP;                                Tmp.x=i,tmp.y=j;                            Q.push (TMP); }                        }                    }                }            }        }    }}intMain () {scanf("%d%d", &n,&m); for(intI=1; i<=n;i++) { for(intj=1; j<=m;j++) {scanf("%d",&MapI [j]);if(MapI [j]==3) St.x=i,st.y=j;Else if(MapI [j]==4) Ed.x=i,ed.y=j; }} floodfill (ST.X,ST.Y); for(intI=1; i<=n;i++) { for(intj=1; j<=m;j++) {if(!MapI [j])        FloodFill (I,J); }    }memset(F,0x3f,sizeof(f)); BFS ();if(f[ed.x][ed.y]==0x3f3f3f3f)printf(" -1\n");Else{printf("%d\n", F[ed.x][ed.y]);printf("%lld\n", Num[ed.x][ed.y]); }}

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Bzoj 1698 [Usaco2007 feb]lilypad Pond Lotus leaf Pond bfs+ Shortest