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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,随意剪枝。用优先队列存一下状态。

还有就是注意浮点数溢出的问题。

#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_backtypedef 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\n",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;}

  

HDU 2128 Tempter of the Bone II BFS