• HDU 2128 Tempter of the Bone II BFS


    状压整张图包括每个点的炸弹有没有被拿,墙壁有没有被炸。用优先队列存一下状态。

    还有就是注意整数数溢出的问题。

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <map>
    #include <set>
    #include <vector>
    #include <string>
    #include <queue>
    #include <deque>
    #include <bitset>
    #include <list>
    #include <cstdlib>
    #include <climits>
    #include <cmath>
    #include <ctime>
    #include <algorithm>
    #include <stack>
    #include <sstream>
    #include <numeric>
    #include <fstream>
    #include <functional>
    
    using namespace std;
    
    #define MP make_pair
    #define PB push_back
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef vector<int> VI;
    typedef pair<int,int> pii;
    const int INF = INT_MAX / 3;
    const double eps = 1e-8;
    const LL LINF = 1e17;
    const double DINF = 1e60;
    const int dx[] = {0,0,1,-1};
    const int dy[] = {1,-1,0,0};
    const int maxn = 8;
    char mp[maxn][maxn];
    int n,m,sx,sy,ex,ey;
    
    
    struct Node {
        int x,y,dis,bomb;
        ULL maze,vis;
        Node(int x,int y,ULL maze,int dis,int bomb,ULL vis): 
            x(x),y(y),maze(maze),dis(dis),bomb(bomb),vis(vis) {}
        bool operator < (const Node &p) const {
            if(dis != p.dis) return dis > p.dis;
            int e_dis1 = abs(x - ex) + abs(y - ey);
            int e_dis2 = abs(p.x - ex) + abs(p.y - ey);
            if(e_dis1 != e_dis2) return e_dis1 > e_dis2;
            if(bomb != p.bomb) return bomb < p.bomb;
            return maze > p.maze;
        }
    };
    
    struct st_Node {
        int x,y,bomb,dis;
        ULL maze,vis;
        st_Node(Node &p) {
            x = p.x; y = p.y; bomb = p.bomb;
            maze = p.maze; vis = p.vis; dis = p.dis;
        }
    
        bool operator < (const st_Node &p) const {
            if(x != p.x) return x < p.x;
            if(y != p.y) return y < p.y;
            if(bomb != p.bomb) return bomb < p.bomb;
            if(maze != p.maze) return maze < p.maze;
            return vis < p.vis;
        }
    };
    
    set<st_Node> st;
    
    
    inline int U(int x,int y) {
        return x * m + y;
    }
    
    inline bool in_bd(int x,int y) {
        return x >= 0 && x < n && y >= 0 && y < m;
    }
    
    inline ULL maze_comp() {
        ULL ret = 0;
        for(int i = 0;i < n;i++) {
            for(int j = 0;j < m;j++) {
                if(mp[i][j] == 'X') ret |= (1LL << U(i,j));
            }
        }
        return ret;
    }
    
    inline bool insert_st(Node &tmp) {
        st_Node now(tmp);
        set<st_Node>::iterator ret = st.find(now);
        if(ret == st.end()) {
            st.insert(now); return true;
        }
        return false;
    }
    
    void printmap(ULL maze,int x,int y) {
        for(int i = 0;i < n;i++) {
            for(int j = 0;j < m;j++) {
                if(i == x && j == y) printf("@");
                else if(maze & (1LL << U(i,j))) printf("X");
                else printf(".");
            }
            puts("");
        }
    }
    
    void solve() {
        st.clear();
        priority_queue<Node> q;
        q.push(Node(sx,sy,maze_comp(),0,0,0));
        int ans = INF;
        while(!q.empty()) {
            Node now = q.top(); q.pop();
            int x = now.x, y = now.y, dis = now.dis, bomb = now.bomb;
            ULL maze = now.maze, vis = now.vis;
            if(x == ex && y == ey) {
                ans = dis;
                break;
            }
            for(int i = 0;i < 4;i++) {
                int nx = x + dx[i], ny = y + dy[i];
                if(!in_bd(nx,ny)) continue;
                if(maze & (1LL << U(nx,ny))) {
                    //如果是墙
                    if(bomb == 0) continue;
                    int nbomb = bomb - 1;
                    ULL nmaze = maze ^ (1LL << U(nx,ny));
                    Node nnode = 
                        Node(nx,ny,nmaze,dis + 2,nbomb,vis);
                    if(insert_st(nnode)) q.push(nnode);
                }
                else {
                    //空地
                    if (mp[nx][ny] >= '1' && mp[nx][ny] <= '9' && !(vis & (1LL << U(nx,ny)))) {
                        //炸弹储备
                        ULL nvis = vis | (1LL << U(nx,ny));
                        int nbomb = bomb + mp[nx][ny] - '0';
                        Node nnode = 
                            Node(nx,ny,maze,dis + 1,nbomb,nvis);
                        if(insert_st(nnode)) {
                            q.push(nnode);
                        }
                    }
                    else {
                        Node nnode = Node(nx,ny,maze,dis + 1,bomb,vis);
                        if(insert_st(nnode)) q.push(nnode);
                    }
                }
            }
        }
        printf("%d
    ",ans == INF ? -1 : ans);
    }
    
    int main() {
        while(scanf("%d%d",&n,&m), n) {
            for(int i = 0;i < n;i++) {
                for(int j = 0;j < m;j++) {
                    scanf(" %c",&mp[i][j]);
                    if(mp[i][j] == 'S') {
                        sx = i; sy = j; mp[i][j] = '.';
                    }
                    if(mp[i][j] == 'D') {
                        ex = i; ey = j; mp[i][j] = '.';
                    }
                }
            }
            solve();
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/rolight/p/3939669.html
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