洛谷-P4011 孤岛营救问题

孤岛营救问题

bfs + 状态压缩

对钥匙的状态进行压缩，然后 bfs 剪枝搜索

#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;
int dp[20][20][1 << 16 | 1];
int dr[20][20][20][20];
int dor[20][20];
const int xi[4] = {0, 0, 1, -1};
const int yi[4] = {1, -1, 0, 0};
int n, m, p;

struct node
{
    int x, y, dis, status;
    node(){}
    node(int _x, int _y, int _dis, int _status){x = _x; y = _y; dis = _dis; status = _status;}
};

inline bool check(int x, int y)
{
    return x >= 1 && x <= n && y >= 1 && y <= m;
}

int bfs()
{
    int ans = n * m * 100;
    queue<node>q;
    q.push({1, 1, 0, dor[1][1]});
    while(q.size())
    {
        node now = q.front();
        q.pop();
        if(now.x == n && now.y == m)
        {
            ans = now.dis < ans ? now.dis : ans;
            continue;
        }
        int& dis = dp[now.x][now.y][now.status];
        if(dis <= now.dis) continue;
        dis = now.dis;
        for(int i=0; i<4; i++)
        {
            int xx = now.x + xi[i];
            int yy = now.y + yi[i];
            if(check(xx, yy))
            {
                int way = dr[now.x][now.y][xx][yy];
                if(way == 0 || (way > 0 && (now.status & (1 << way)) == 0)) continue;
                q.push(node(xx, yy, dis + 1, now.status | dor[xx][yy]));
            }
        }
    }
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m >> p;
    for(int i=0; i<=n; i++)
        for(int j=0; j<=m; j++)
            for(int u=0; u<=n; u++)
                for(int v=0; v<=m; v++)
                    dr[i][j][u][v] = -1;
    int temp = n * m * 100;
    for(int i=0; i<=n; i++)
        for(int j=0; j<=m; j++)
            for(int u=1<<(p+2); u>=0; u--)
                dp[i][j][u] = temp;
    int k;
    cin >> k;
    while(k--)
    {
        int ax, ay, bx, by, x;
        cin >> ax >> ay >> bx >> by >> x;
        dr[ax][ay][bx][by] = x;
        dr[bx][by][ax][ay] = x;
    }
    cin >> k;
    while(k--)
    {
        int x, y, z;
        cin >> x >> y >> z;
        dor[x][y] |= 1 << z;
    }
    int ans = bfs();
    if(ans == temp)
        ans = -1;
    cout << ans << endl;
    return 0;
}

标签：P4011,洛谷,int,孤岛,&&,ay,ax,now,bx
来源： https://www.cnblogs.com/dgsvygd/p/16366468.html