AOJ.866 飞越原野 (三维BFS)

AOJ.866 飞越原野 (三维BFS)

题意分析

点我挑战题目

相比于普通的BFS，要多一维来记录当前剩余的体力。而且还要额外的一层循环来处理，飞过的路程。

代码总览

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#define INF 0x3f3f3f3f
#define nmax 200
#define MEM(x) memset(x,0,sizeof(x))
using namespace std;
bool visit[nmax][nmax][nmax],isfind =  false;
char mp[nmax][nmax];
struct nod{
    int x,y,e;
}node[nmax];
int m,n,o,ans;
int spx[] = {0,1,0,-1};
int spy[] = {1,0,-1,0};
bool check(int x, int y, int e)
{
    if(x<0 || x >= m || y<0 || y>=n || mp[x][y] =='L' || visit[x][y][e] == true)
        return false;
    else
        return true;
}
void bfs()
{
    nod temp;
    temp.x = 0; temp.y = 0; temp.e = o;
    visit[temp.x][temp.y][temp.e] = true;
    queue<nod> q;
    if(!q.empty()) q.pop();
    q.push(temp);
    ans = 0;
    while(!q.empty()){
        int nowsize = q.size();
        while(nowsize--){
            nod tep = temp = q.front();
            q.pop();
            // get final or not
            if(tep.x == m-1 && tep.y == n-1){
                isfind = true;
                break;
            }
            for(int i = 0 ;i<4;++i){
                tep.x =temp.x + spx[i];
                tep.y =temp.y + spy[i];
                if(check(tep.x,tep.y,tep.e)){
                    visit[tep.x][tep.y][tep.e] = true;
                    q.push(tep);
                }
            }
            for(int i = 0; i<4;++i){
                for(int j = 1; j<=temp.e;++j){
                    tep.x = temp.x + spx[i] *j;
                    tep.y = temp.y + spy[i] *j;
                    tep.e = temp.e - j;
                    if(check(tep.x,tep.y,tep.e)){
                        visit[tep.x][tep.y][tep.e] = true;
                        q.push(tep);
                    }
                }
            }
        }
        if(isfind) break;
        ans++;
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d %d %d",&m,&n,&o) != EOF){
        MEM(visit);
        isfind  = false;
        for(int i = 0; i<m;++i)
            scanf("%s",mp[i]);
        bfs();
        if(isfind) printf("%d\n",ans);
        else printf("impossible\n");
    }
    return 0;
}