藏宝图

蒜头君得到一张藏宝图。藏宝图是一个 $10\times 10$的方格地图，图上一共有 $10$ 个宝藏。有些方格地形太凶险，不能进入。

整个图只有一个地方可以出入，即是入口也是出口。蒜头君是一个贪心的人，他规划要获得所有宝藏以后才从出口离开。

藏宝图上从一个方格到相邻的上下左右的方格需要 1

$1$ 天的时间，蒜头君从入口出发，找到所有宝藏以后，回到出口，最少需要多少天。

玩命的搜索啊

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <windows.h>
using namespace std;
bool book[10][10];
int bz[11][2] = {0,7,1,6,2,4,3,1,3,8,4,4,6,8,7,6,8,1,9,6,0,0};//每个宝藏的坐标
int Map[10][10]= //整个地图 2表示不能走 1表示宝藏
{
{0,0,0,0,0,0,0,1,0,0},
{0,0,0,2,0,0,1,0,0,0},
{0,2,0,0,1,0,0,2,0,0},
{0,1,0,0,0,2,0,0,1,0},
{0,2,0,0,1,0,2,0,0,0},
{0,0,2,0,0,0,0,2,0,0},
{0,0,0,0,0,2,0,0,1,0},
{0,2,0,2,0,0,1,0,0,0},
{0,1,0,0,0,0,2,2,0,0},
{0,0,2,0,0,2,1,0,0,0}
};
struct node
{
int x,y,step;
} Now,Next;
int dir[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
int bfs(int sx,int sy, int ex, int ey)
{
memset(book,0,sizeof(book));
Now.x = sx;
Now.y = sy;
Now.step = 0;
queue<node> q;
q.push(Now);
book[sx][sy] = true;
while(!q.empty())
{
Now = q.front();
q.pop();
if(Now.x == ex && Now.y == ey)
{
return Now.step;
}
for(int i = 0; i < 4; i++)
{
Next.x = Now.x + dir[i][0];
Next.y = Now.y + dir[i][1];
Next.step = Now.step + 1;
if(Next.x >= 0 && Next.x <= 9 && Next.y >= 0 && Next.y <= 9 && Map[Next.x][Next.y] != 2 && book[Next.x][Next.y] == false)
{
book[Next.x][Next.y] = true;
q.push(Next);
}
}
}
return -1;
}
int main()
{
int minn = 999999;
int ans[11][11];
for(int i = 0; i < 11; i++)
{
for(int j = 0; j < 11; j++)
{
ans[i][j] = bfs(bz[i][0],bz[i][1],bz[j][0],bz[j][1]);
}
}
int num[10] = {0,1,2,3,4,5,6,7,8,9};
do
{
int sum = 0;
sum += ans[10][num[0]];
for(int i = 0; i < 9; i++)
{
sum += ans[num[i]][num[i+1]];
}
sum += ans[num[9]][10];
minn = min(minn,sum);
}
while(next_permutation(num,num+10));
cout << minn << endl;
return 0;
}

最后

以上就是活泼纸飞机最近收集整理的关于藏宝图的全部内容，更多相关藏宝图内容请搜索靠谱客的其他文章。