Problem Description

FJ has installed a beautiful pond for his cows' aesthetic enjoyment and exercise.

The rectangular pond has been partitioned into square cells of M rows and N columns (1 ≤ M ≤ 30; 1 ≤ N ≤ 30). Some of the cells have astonishingly sturdy lilypads; others have rocks; the remainder are just beautiful, cool, blue water.
Bessie is practicing her ballet moves by jumping from one lilypad to another and is currently located at one of the lilypads. She wants to travel to another lilypad in the pond by jumping from one lilypad to another.
Surprising only to the uninitiated, Bessie's jumps between lilypads always appear as a chess-knight's move: one move in one direction and then two more in the orthogonal direction (or perhaps two in one direction and then one in the orthogonal direction).
Farmer John is observing Bessie's ballet drill and realizes that sometimes she might not be able to jump to her destination lilypad because intermediary lilypads are missing.
Ever thrifty, he wants to place additional lilypads so she can complete her quest (perhaps quickly, perhaps by using a large number of intermediate lilypads). Of course, lilypads cannot be placed where rocks already intrude on a cell.
Help Farmer John determine the minimum number of additional lilypads he has to place, and in how many ways he can place that minimum number.

Input

Line 1: Two space-separated integers: M and N?
Lines 2..M+1: Line i+1 describes row i of the pond using N space-separated integers with these values: 0 indicates empty water; 1 indicates a lilypad in place; 2 indicates rock in place; 3 indicates the lilypad Bessie starts on; 4 indicates the lilypad Bessie wants to travel to.

Output

Line 1: One integer: the minimum number of additional lilypads required. If it is not possible to help Bessie jump to her destination, print -1.?
Line 2: One integer: the total number of possible ways the additional lilypads can be positioned. This number is guaranteed to fit into a single 64-bit signed integer. Do not output this line if line 1 contains -1.

Sample Input

4 5

1 0 0 0 0
3 0 0 0 0
0 0 2 0 0
0 0 0 4 0

Sample Output

2

3

題意：給出一 M * N 的矩陣，0為空，1為荷葉，2有石頭，3是起點，4是終點，牛要從起點跳到終點，跳時只能走象棋中的馬步，要幫助牛從起點到終點，最少放多少片荷葉，以及放最少荷葉的最小方案數。

思路：先把能到達的每個需要建荷花的點進行預處理，將中間跳躍的過程省略，這樣從起點到終點的最短路就是最少要放的荷花數，最后的方案數為上一點的方案數相加。

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<set> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 51 #define MOD 123 #define E 1e-6 using namespace std; int m,n; int dir[8][2]= {{1,2},{1,-2},{2,1},{2,-1},{-1,2},{-1,-2},{-2,1},{-2,-1}}; int g[N][N]; long long dis[N][N],ans[N][N]; int vis[N][N]; int d[N][N][N][N]; long long len[N][N]; int judge(int x,int y) {if(x>=1&&x<=m&&y>=1&&y<=n)return 1;return 0; } void bfs1(int x0,int y0) {queue <int> queue_x;queue <int> queue_y;queue_x.push(x0);queue_y.push(y0);memset(vis,0,sizeof(vis));vis[x0][y0]=1;while(!queue_x.empty()) {int dirx=queue_x.front();int diry=queue_y.front();queue_x.pop();queue_y.pop();for(int i=0; i<8; i++) {int xx=dirx+dir[i][0];int yy=diry+dir[i][1];if(judge(xx,yy)) {if(vis[xx][yy]==0) {vis[xx][yy]=1;if(g[xx][yy]==1) {queue_x.push(xx);queue_y.push(yy);} elsed[x0][y0][xx][yy]=1;}}}} } void bfs2(int x0,int y0) {queue <int> queue_x;queue <int> queue_y;queue_x.push(x0);queue_y.push(y0);memset(ans,0,sizeof(ans));memset(dis,INF,sizeof(dis));ans[x0][y0]=1;dis[x0][y0]=0;len[x0][y0]=0;while(!queue_x.empty()) {int dirx=queue_x.front();int diry=queue_y.front();queue_x.pop();queue_y.pop();if(g[dirx][diry]==2)continue;for(int i=1; i<=m; i++) {for(int j=1; j<=n; j++) {if(d[dirx][diry][i][j]) {if(dis[i][j]==dis[dirx][diry]+1)ans[i][j]+=ans[dirx][diry];else if(dis[i][j]>dis[dirx][diry]+1) {dis[i][j]=dis[dirx][diry]+1;ans[i][j]=ans[dirx][diry];queue_x.push(i);queue_y.push(j);}}}}} } int main() {int start_x,start_y,end_x,end_y;while(scanf("%d%d",&m,&n)!=EOF) {for(int i=1; i<=m; i++) {for(int j=1; j<=n; j++) {scanf("%d",&g[i][j]);if(g[i][j]==3) { //記錄始點start_x=i;start_y=j;}if(g[i][j]==4) { //記錄終點end_x=i;end_y=j;}}}memset(d,0,sizeof(d));for(int i=1; i<=m; i++) //預處理for(int j=1; j<=n; j++)bfs1(i,j);bfs2(start_x,start_y);if(dis[end_x][end_y]>INF)printf("-1\n");elseprintf("%lld\n%lld\n",dis[end_x][end_y]-1,ans[end_x][end_y]);}return 0; }

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