HDU-1372Knight Moves(BFS)

BFS模板题

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1372

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思路：
这题就思路来说，DFS和BFS都可以得到最优解，不过dfs会生成大量重复非最优解，即使优化（用一个二维数组保存到每格的最短时间）也会超时。

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下面先附上dfs代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
int sx, sy, ex, ey, ans = 64;
int vis[10][10];
int mov[8][2] = {{-2,+1},{-1,+2},{+1,+2},{+2,+1},{+2,-1},{+1,-2},{-1,-2},{-2,-1}};

bool get(int x, int y){
    if (x == ex&&y == ey) return true;
    return false;
}

bool over(int x, int y){
    if (x<0||x>=8||y<0||y>=8) return true;
    return false;
}

void dfs(int cur, int x, int y){
    if (cur >= ans) return;
    if (!vis[x][y]) vis[x][y] = cur;
    if (get(x, y)){
        ans = ans<cur?ans:cur;
        return;
    }
    for (int i = 0; i < 8; i++){
        if (!over(x+mov[i][0], y+mov[i][1])&&cur<=vis[x][y]){
            vis[x][y] = cur;
            dfs(cur+1, x+mov[i][0], y+mov[i][1]);
        }
    }
}

int main()
{
    freopen("E://in.txt", "r", stdin);
    char s1, s2, mid;
    while(scanf("%c%d%c%c%d", &s1, &sy, &mid, &s2, &ey) != EOF){
        getchar();
        sy -= 1;
        ey -= 1;
        sx = s1 - 'a';
        ex = s2 - 'a';
        dfs(0, sx, sy);
        printf("To get from %c%d to %c%d takes %d knight moves.\n",s1, sy+1, s2, ey+1, ans);
        ans = 64;
        memset(vis, 0, sizeof(vis));
    }

    return 0;
}

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用BFS就是一道很简单的模板题了
下面是AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;

int sx, sy, ex, ey, ans;
int vis[10][10];
int mov[8][2] = {{-2,+1},{-1,+2},{+1,+2},{+2,+1},{+2,-1},{+1,-2},{-1,-2},{-2,-1}};

typedef struct point{
    int cur;
    int x;
    int y;
}point;

bool get(int x, int y){
    if (x == ex&&y == ey) return true;
    return false;
}

bool over(int x, int y){
    if (x<0||x>=8||y<0||y>=8) return true;
    return false;
}

int bfs(){
    queue<point>q;
    point sta = {0, sx, sy};
    q.push(sta);
    while(!q.empty()){
        point now = q.front();
        q.pop();
        if (get(now.x, now.y)){
            ans = now.cur;
            return 0;
        }
        for (int i = 0; i < 8; i++){
            int nextx = now.x + mov[i][0], nexty = now.y + mov[i][1];
            if (!over(nextx, nexty)&&!vis[nextx][nexty]){
                point newp = {now.cur+1, nextx, nexty};
                vis[nextx][nexty] = 1;
                q.push(newp);
            }
        }
    }
}

int main()
{
    //freopen("E://in.txt", "r", stdin);
    char s1, s2, mid;
    while(scanf("%c%d%c%c%d", &s1, &sy, &mid, &s2, &ey) != EOF){
        getchar();
        sy -= 1;
        ey -= 1;
        sx = s1 - 'a';
        ex = s2 - 'a';
        bfs();
        printf("To get from %c%d to %c%d takes %d knight moves.\n",s1, sy+1, s2, ey+1, ans);
        memset(vis, 0, sizeof(vis));
        ans = 0;
    }

    return 0;
}