POJ 3170 Knights of Ni （暴力，双向BFS）

题意：一个人要从2先走到4再走到3，计算最少路径。

析：其实这个题很水的，就是要注意，在没有到4之前是不能经过3的，一点要注意。其他的就比较简单了，就是一个双向BFS，先从2搜到4，再从3到搜到4，

然后求最短路即可。

代码如下：

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <string>

#include <algorithm>

#include <vector>

#include <map>

using namespace std ;

typedef long long LL;

typedef pair<int, int> P;

template<class T>T scan(T &x){

    int f = 1;  x = 0;  char c;

    while(c = getchar(), c < 48)  if(c == '-')  f = -1;

    do  x = x * 10 + (c^48);

    while(c = getchar(), c > 47);

    x *= f;

    return x;

}

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f3f;

const double eps = 1e-11;

const int maxn = 1000 + 5;

const int dr[] = {0, 0, -1, 1};

const int dc[] = {-1, 1, 0, 0};

struct node{

    int x, y;

    node(int xx = 0, int yy = 0) : x(xx), y(yy) { }

};

int n, m;

int a[maxn][maxn];

int vis1[maxn][maxn];

int vis2[maxn][maxn];

vector<node> v;

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

void bfs(int r, int c, int vis[][maxn], bool ok){

    queue<node> q;

    q.push(node(r, c));

    vis[r][c] = 0;

    while(!q.empty()){

        node u = q.front();  q.pop();

        for(int i = 0; i < 4; ++i){

            int x = u.x + dr[i];

            int y = u.y + dc[i];

            if(vis[x][y] != INF || !is_in(x, y) || a[x][y] == 1)  continue;

            if(!ok && a[x][y] == 3)  continue;

            vis[x][y] = vis[u.x][u.y] + 1;

            q.push(node(x, y));

        }

    }

}

int main(){

    scanf("%d %d", &m, &n);

    int sx, sy, ex, ey;

    for(int i = 0; i < n; ++i)

        for(int j = 0; j < m; ++j){

            scanf("%d", &a[i][j]);

            if(a[i][j] == 4) v.push_back(node(i, j));

            else if(a[i][j] == 2)  sx = i, sy = j;

            else if(a[i][j] == 3)  ex = i, ey = j;

        }

    for(int i = 0; i < n; ++i)

        for(int j = 0; j < m; ++j)

            vis1[i][j] = vis2[i][j] = INF;

    bfs(sx, sy, vis1, false);

    bfs(ex, ey, vis2, true);

    int ans = INF;

    for(int i = 0; i < v.size(); ++i){

        ans = min(ans, vis1[v[i].x][v[i].y]+vis2[v[i].x][v[i].y]);

    }

    cout << ans << endl;

    return 0;

}