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HDU T1253 胜利大逃亡
题解:
这题并不难,别被题目吓住了,只要搞清楚三维坐标怎么摆放,这题就OK了.
因为(0,0,0)在左上后方(造词---),所以坐标抽该这么摆
对应的关系是:
A——X
B——Y
C——Z
这关系在我代码中对maps数组的使用就可以看出来。
代码:
#include<cstdio>
#include<iostream>
#include<queue>
#define maxn 55
using namespace std;
int a,b,c,tl;
int maps[maxn][maxn][maxn];
int mov[6][3] = {0,-1,0, 0,1,0, 0,0,-1, 0,0,1, 1,0,0, -1,0,0};
struct Node{
int x,y,z,st;
}now,nex;
bool charge(int x,int y,int z){
if(x < 0 || x >= a|| y < 0|| y >= b|| z < 0|| z >= c || maps[x][y][z])
return false;
return true;
}
int Bfs(int st){
now.x = 0;
now.y = 0;
now.z = 0;
now.st = st;
queue<Node> Q;
Q.push(now);
while(!Q.empty()){
now = Q.front();
Q.pop();
if(now.x == a-1 && now.y == b-1 && now.z == c-1 && now.st <= tl)
return now.st;
for(int i = 0; i < 6; ++i){
nex.x = now.x + mov[i][0];
nex.y = now.y + mov[i][1];
nex.z = now.z + mov[i][2];
if(charge(nex.x,nex.y,nex.z)){
nex.st = now.st + 1;
maps[nex.x][nex.y][nex.z] = 1;
Q.push(nex);
}
}
}
return -1;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d%d%d",&a,&b,&c,&tl);
for(int i = 0; i < a; ++i)
for(int j = 0; j < b; ++j)
for(int k = 0; k < c; ++k)
scanf("%d",&maps[i][j][k]);
printf("%d\n",Bfs(0));
}
}