FZU 2124 吃豆人 bfs

题目链接：吃豆人

比赛的时候写的bfs，纠结要不要有vis数组设置已被访问，没有的话死循环，有的话就不一定是最优解了。【此时先到的不一定就是时间最短的。】于是换dfs，WA。

赛后写了个炒鸡聪明的dfs，TLE，才发现时间复杂度好像是4^(n*m)。T_T

依然感觉这个dfs很棒。

bfs已AC，怎么解决的这个问题呢，如果当前位置next 被优化了则加入队列，以此优化其他位置，否则不加入队列。T_T好有道理~~~

感觉bfs和dfs好神奇的说~

dfs TLE代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <cmath>
#include <queue>
#define inf 100000000
using namespace std;

char mp[30][30];

struct Node{
    int x, y;
}st, ed, tool, temp, nxt;

int vis[30][30];
double step[30][30];
int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
int n, m;

bool check(Node temp) {
    if (temp.x >= 0 && temp.x < n && temp.y >= 0 && temp.y < m && !vis[temp.x][temp.y] && mp[temp.x][temp.y] != 'X') {
        return true;
   )}
    return false;
}

void get(Node temp, Node ed) {
    if (temp.x == ed.x) {
        int minn = min(temp.y, ed.y);
        int maxm = max(temp.y, ed.y);
        for (int i=minn+1; i<maxm; ++i) {
            if (mp[ed.x][i] == 'X' || mp[ed.x][i] == 'S') return;
        }
        int t = abs(temp.y - ed.y);
        step[ed.x][ed.y] = min(step[temp.x][temp.y]+t*0.2,step[ed.x][ed.y]);
    }
    else if (temp.y == ed.y) {
        int minn = min(temp.x, ed.x);
        int maxm = max(temp.x, ed.x);
        for (int i=minn+1; i<maxm; ++i) {
            if (mp[i][ed.y] == 'X' || mp[i][ed.x] == 'S') return;
        }
        int t = abs(temp.x - ed.x);
        step[ed.x][ed.y] = min(step[temp.x][temp.y]+t*0.2, step[ed.x][ed.y]);
    }
}

void dfs(Node st, Node ed, bool v) {
    for (int i=0; i<4; ++i) {
    nxt.x = st.x + dir[i][0];
    nxt.y = st.y + dir[i][1];
    if (check(nxt)) {
        double t = step[st.x][st.y];
        if (v) t += 0.5;
            else t += 1;
        step[nxt.x][nxt.y] = min(step[nxt.x][nxt.y], t);
        vis[nxt.x][nxt.y] = 1;
        get(nxt, ed);
        if (nxt.x == tool.x && nxt.y == tool.y) dfs(nxt, ed, true);
        else dfs(nxt, ed, v);
      }
    }
   vis[st.x][st.y] = 0;
   return;
}

int main() {
    while(~scanf("%d%d", &n, &m)) {
        bool v = false;
        memset(vis, 0, sizeof(vis));
        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                step[i][j] = inf;
            }
        }
        getchar();
        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                scanf("%c", &mp[i][j]);
                if (mp[i][j] == 'P') {
                    st.x = i, st.y = j;
                }
                else if (mp[i][j] == 'B') {
                    ed.x = i, ed.y = j;
                }
                else if (mp[i][j] == 'S') {
                    tool.x = i, tool.y = j;
                }
            }
            if (i != n-1) scanf("
");
        }
        step[st.x][st.y] = 0;
        vis[st.x][st.y] = 1;
        dfs(st, ed, v);
        get(st, ed);
        double ans = step[ed.x][ed.y];
        if (ans != inf)
            printf("%.1lf
", ans);
       else printf("-1
");
    }
    return 0;
}

bfs AC 代码：

/*
直接一个bfs，用两个数组保存有工具和没有工具时需要的时间。
*/

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#define inf 1000000000
using namespace std;

char mp[30][30];
int n, m;

struct Node {
    int x, y, s; // 0 表示没有工具 1 表示有工具了。
}st, ed, temp, now, nxt;

int step[30][30][2];

int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};

void bfs(Node st) {
    queue<Node> que;
    que.push(st);
    step[st.x][st.y][0] = 0;
    while(!que.empty()) {
        now = que.front();
        que.pop();
        if (now.x == ed.x) {
            int miny = min(now.y, ed.y);
            int maxy = max(now.y, ed.y);
            bool get = true;
            for (int y=miny+1; y<maxy; ++y) {
                if (mp[now.x][y] == 'X' || mp[now.x][y] == 'S') {
                        get = false;
                        break;
                }
            }
            if (get) step[ed.x][ed.y][now.s] = min(step[now.x][now.y][now.s] + 2 * (maxy - miny), step[ed.x][ed.y][now.s]);
        }

        if (now.y == ed.y) {
            int minx = min(now.x, ed.x);
            int maxx = max(now.x, ed.x);
            bool get = true;
            for (int x=minx+1; x<maxx; ++x) {
                if (mp[x][ed.y] == 'X' || mp[x][ed.y] == 'S') {
                    get = false;
                }
            }
            if (get) step[ed.x][ed.y][now.s] = min(step[now.x][now.y][now.s]  + 2 * (maxx - minx), step[ed.x][ed.y][now.s]);
        }

        int T;
        for (int i=0; i<4; ++i) {
            nxt.x = now.x + dir[i][0];
            nxt.y = now.y + dir[i][1];
            nxt.s = now.s;
            if (nxt.x<0 || nxt.y<0 || nxt.x>=n || nxt.y>=m) continue;
            if (mp[nxt.x][nxt.y] == 'X') continue;
            if (now.s) T = 5;
            else T = 10;
            if (step[nxt.x][nxt.y][nxt.s] > step[now.x][now.y][now.s] + T) {
                step[nxt.x][nxt.y][nxt.s] = step[now.x][now.y][now.s] + T;
                if (mp[nxt.x][nxt.y] == 'S') {
                    nxt.s = 1;
                    step[nxt.x][nxt.y][1] = step[nxt.x][nxt.y][0];
                }
                que.push(nxt);
            }
        }
    }
}




int main() {
    while(cin >> n >> m) {
        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                for (int k=0; k<2; ++k) {
                    step[i][j][k] = inf;
                }
            }
        }

        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                cin >> mp[i][j];
                if (mp[i][j] == 'P') {
                    st.x = i, st.y = j;
                }
                else if (mp[i][j] == 'B') {
                    ed.x = i, ed.y = j;
                }
            }
        }

        st.s = 0;
        bfs(st);
        int ans = min(step[ed.x][ed.y][0], step[ed.x][ed.y][1]);
        if (ans == inf) cout << "-1
";
        else printf("%.1lf
", ans*1.0/10);
    }
    return 0;
}