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小鼠迷宫问题

题面

输入

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8 8 3
3 3
4 5
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2 1
7 7

输出

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2
11
96

思路

bfs搜最短路
dfs最短路数量

只用一种应该也可以 bfs 分步乘法?

ac代码

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#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<unordered_map>
#include<queue>

using namespace std;

const int N = 1000;

typedef pair<int, int> PII;

int n, m, k, ax, ay, bx, by,ans,t,step;
int map[N][N];
bool st[N][N];
queue<PII> q;
int dist[N][N];
int dx[4] = { 1,-1,0,0 }, dy[4] = { 0,0,1,-1 };

int bfs()
{
q.push({ ax,ay });
dist[ax][ay] = 0;
st[ax][ay] = 0;

while (!q.empty())
{
PII t = q.front();
q.pop();

int x = t.first, y = t.second;

if (x == bx && y == by) return dist[x][y];

for (int i = 0; i <= 3; i++)
{
int xx = x + dx[i], yy = y + dy[i];
if (xx >= 1 && xx <= n && yy >= 1 && yy <= n && !st[xx][yy]&&!map[xx][yy])
{
q.push({ xx,yy });
dist[xx][yy] = dist[x][y] + 1;
}
}
}

return -1;
}

void dfs(int x,int y,int s)
{
if (s > t)return;
if (abs(bx - x) + abs(by - y) + s > t)return;
if (x == bx && y == by && s==t)
{
step+=1;
return;
}

for (int i = 0; i <= 3; i++)
{
int xx = x + dx[i], yy = y + dy[i];
if (xx >= 1 && xx <= n && yy >= 1 && yy <= n && !st[xx][yy] && !map[xx][yy])
{
st[xx][yy] = true;
dfs(xx, yy, s + 1);
st[xx][yy] = false;
}
}
}

int main()
{
while (cin >> n >> m >> k)
{
ans = 0x3f3f3f3f; step = 0;
memset(map, 0, sizeof map);
memset(st, false, sizeof st);
while (k--)
{
int i, j;
cin >> i >> j;
map[i][j] = 1;
}

cin >> ax >> ay >> bx >> by;

t = bfs();
if (t == -1) {
cout << "No Solution!" << endl<<"0"<<endl;
}
else {
memset(st, false, sizeof st);
dfs(ax, ay, 0);
cout << t << endl<<step<<endl;
}


}
}
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