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分享刘汝佳大神的代码，英文是刘汝佳大神的注释，中文是我自己加的注释

Posted by sapphirebitter at 2017-08-31 17:25:02 on Problem 3523
```#include<cstdio>
#include<cstring>
#include<cctype>
#include<queue>
using namespace std;

const int maxs = 20;
const int maxn = 150; /// 75% cells plus 2 fake nodes
const int dx[]= {1,-1,0,0,0}; /// 4 moves, plus "no move"
const int dy[]= {0,0,1,-1,0};

inline int ID(int a, int b, int c)
{
return (a<<16)|(b<<8)|c;///由于int 类型保存8个字节，所以这个操作将a放到了最前面的8个字节，b放到了中间的8个字节，c放到了最后的8个字节
}

int s[3], t[3]; /// starting/ending position of each ghost

int deg[maxn], G[maxn][5]; /// target cells for each move (including "no move").deg 表示出度，存储能够前进的方向

inline bool conflict(int a, int b, int a2, int b2)
{
return a2 == b2 || (a2 == b && b2 == a);///第一个表示a,b进入同一空格，第二个表示2者交换位置
}

int d[maxn][maxn][maxn]; /// distance from starting state

int bfs()
{
queue<int> q;
memset(d, -1, sizeof(d));
q.push(ID(s[0], s[1], s[2])); /// starting node
d[s[0]][s[1]][s[2]] = 0;
while(!q.empty())
{
int u = q.front();
q.pop();
int a = u>>16, b = (u>>8)&0xff, c = u&0xff;///&0xff表示取最后的8个字节
if(a == t[0] && b == t[1] && c == t[2]) return d[a][b][c]; /// solution found
for(int i = 0; i < deg[a]; i++)
{
int a2 = G[a][i];
for(int j = 0; j < deg[b]; j++)
{
int b2 = G[b][j];
if(conflict(a, b, a2, b2)) continue;
for(int k = 0; k < deg[c]; k++)
{
int c2 = G[c][k];
if(conflict(a, c, a2, c2)) continue;
if(conflict(b, c, b2, c2)) continue;
if(d[a2][b2][c2] != -1) continue;///判断是否被访问过
d[a2][b2][c2] = d[a][b][c]+1;
q.push(ID(a2, b2, c2));
}
}
}
}
return -1;
}

int main()
{
int w, h, n;

while(scanf("%d%d%d\n", &w, &h, &n) == 3 && n)
{
char maze[20][20];
for(int i = 0; i < h; i++)
fgets(maze[i], 20, stdin);

/// extract empty cells
int cnt, x[maxn], y[maxn], id[maxs][maxs]; ///cnt is the number of empty cells
cnt = 0;
for(int i = 0; i < h; i++)
for(int j = 0; j < w; j++)
if(maze[i][j] != '#')
{
x[cnt] = i;
y[cnt] = j;
id[i][j] = cnt;
if(islower(maze[i][j])) s[maze[i][j] - 'a'] = cnt;
else if(isupper(maze[i][j])) t[maze[i][j] - 'A'] = cnt;
cnt++;
}

/// build a graph of empty cells
for(int i = 0; i < cnt; i++)
{
deg[i] = 0;
for(int dir = 0; dir < 5; dir++)
{
int nx = x[i]+dx[dir], ny = y[i]+dy[dir];
/// "Outermost cells of a map are walls" means we don't need to check out-of-bound
if(maze[nx][ny] != '#') G[i][deg[i]++] = id[nx][ny];///出度+1，指向空格区域
}
}

/// add fakes nodes so that in each case we have 3 ghosts. this makes the code shorter
if(n <= 2)
{
deg[cnt] = 1;
G[cnt][0] = cnt;
s[2] = t[2] = cnt++;
}
if(n <= 1)
{
deg[cnt] = 1;
G[cnt][0] = cnt;
s[1] = t[1] = cnt++;
}

printf("%d\n", bfs());
}
return 0;
}```

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