Uva 816 Abbott's Revenge 紫书165页例题

最新推荐文章于 2022-01-11 14:10:33 发布

Cwolf9

最新推荐文章于 2022-01-11 14:10:33 发布

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分类专栏： DFS相关和BFS等 ACM算法之旅

本文链接：https://blog.youkuaiyun.com/qq_39599067/article/details/84721760

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本文深入探讨了使用C++实现的一种迷宫寻路算法，通过定义方向、状态转移和路径回溯等关键概念，详细介绍了如何利用队列和地图数据结构来寻找从起点到终点的最短路径。文章提供了完整的代码示例，包括状态转移矩阵、路径记录和读取案例等功能。

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#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;

const int mod = 1000000007;
const int INF = 0x3f3f3f3f;
int dir[4][2]={-1,0,0,1,0,-1,1,0};//上N右E左W下S
int to[4][3]={//FLR
    {0,2,1},
    {1,0,3},
    {2,3,0},
    {3,1,2}
};
int n, m;
int sx, sy, ex, ey;
char SIN;
map<char, int> mp;
int buf[10][10][4][3];
int vis[10][10][4];
int fa[1000];
struct lp {
    int x, y, step, f;
};
bool readcase() {
    string s; cin>>s;
    if(s == "END") return 0;
    cout<<s<<"\n";
    memset(buf, 0, sizeof(buf)); memset(vis, 0x3f, sizeof(vis));
    cin>>sx>>sy;
    cin>>SIN;
    cin>>ex>>ey;
    cin>>n;
    while(n != 0) {
        cin>>m;cin>>s;
        while(s != "*") {
            int len = s.length();
            for(int i = 1; i < len; ++i) {
                buf[n][m][mp[s[0]]][mp[s[i]]] = 1;
            }
            cin>>s;
        }
        cin>>n;
    }
    return 1;
}
void solve() {
    int ax = sx + dir[mp[SIN]][0], ay = sy + dir[mp[SIN]][1];
    queue<lp> Q;
    fa[ax*10+ay+mp[SIN]*100]=sx*10+sy;
    Q.push(lp{ax,ay,1,mp[SIN]});
    int ans = INF, last = -1;
    while(!Q.empty()) {
        lp A = Q.front(); Q.pop();
        if(A.x == ex && A.y == ey) {
            if(A.step < ans) {
                ans = A.step;
                last = A.x*10+A.y+A.f*100;
            }
        }
        for(int i = 0; i < 3; ++i) {
            if(buf[A.x][A.y][A.f][i] == 0) continue;
            int px = dir[to[A.f][i]][0]+A.x;
            int py = dir[to[A.f][i]][1]+A.y;
            int pf = to[A.f][i];
            int pt = A.step + 1;
            if(px < 1 || px > 9 || py < 1 || py > 9) continue;
            if(vis[px][py][pf] > pt) {
                vis[px][py][pf] = pt;
                fa[px*10+py+pf*100]=A.x*10+A.y+A.f*100;
                Q.push(lp{px,py,pt,pf});
            }
        }
    }
    if(ans == INF) printf("  No Solution Possible\n");
    else {
        std::vector<pair<int,int> > vs;
        while((last%100)/10!=ax||last%10!=ay||last/100!=mp[SIN]) {
            vs.push_back({(last%100)/10,last%10});
            last = fa[last];
        }
        vs.push_back({ax,ay});
        vs.push_back({sx,sy});
        int len = vs.size();
        printf("  ");
        for(int i = len - 1, j = 1; i >= 0; --i, ++j) {
            printf("(%d,%d)", vs[i].first, vs[i].second);
            if(i == 0 || j % 10 == 0) {
                printf("\n");
                if(i != 0) printf("  ");
            }else printf(" ");
        }
    }
}
int main() {
    mp['N']=0;mp['E']=1;mp['W']=2;mp['S']=3;
    mp['F']=0;mp['L']=1;mp['R']=2;
    while(readcase()) solve();
    return 0;
}