题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4771
题面：
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Stealing Harry Potter‘s PreciousTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2207 Accepted Submission(s): 1035 Problem Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon‘s home. But he can‘t bring his precious with him. As you know,
Some rooms are indestructible and some rooms are vulnerable. Goblins always care more about their own safety than their customers‘ properties, so they live in the indestructible rooms and put customers‘ properties in vulnerable rooms. Harry Potter‘s precious Input There are several test cases. In each test cases: The first line are two integers N and M, meaning that the bank is a N × M grid(0<N,M <= 100). Then a N×M matrix follows. Each element is a letter standing for a room. ‘#‘ means a indestructible room, ‘.‘ means a vulnerable room, and the only ‘@‘ means the vulnerable room from which Dudely starts to move. The next line is an integer K ( 0 < K <= 4), indicating there are K Harry Potter‘s precious in the bank. In next K lines, each line describes the position of a Harry Potter‘s precious by two integers X and Y, meaning that there is a precious in room (X,Y). The input ends with N = 0 and M = 0 Output For each test case, print the minimum number of steps Dudely must take. If Dudely can‘t get all Harry‘s things, print -1. Sample Input 2 3 ##@ #.# 1 2 2 4 4 #@## .... #### .... 2 2 1 2 4 0 0 Sample Output -1 5 Source |
解题：
首先找出，每两个点间的最短距离，然后由起点出发，算出由该点到达所有点的最短距离。写的比较复杂，可以稍加改进。
总结：
深搜结束后，一定要注意状态的还原。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
char map[105][105];
bool visit[105][105],vis[5];
int n,m,q,fmx,fmy,tox,toy,x,y,len,tmp,ans,res;
int dis[5][5];
int desx[5],desy[5];
struct node
{
int x,y,step;
};
queue <node> qe;
int bfs()
{
while(!qe.empty())
qe.pop();
node tmp;
tmp.x=fmx;
tmp.y=fmy;
tmp.step=0;
qe.push(tmp);
memset(visit,-1,sizeof(visit));
for(int i=0;i<=m+1;i++)
{
visit[0][i]=visit[n+1][i]=0;
}
for(int i=1;i<=n;i++)
{
visit[i][0]=visit[i][m+1]=0;
}
while(!qe.empty())
{
tmp=qe.front();
qe.pop();
x=tmp.x;
y=tmp.y;
len=tmp.step;
if(map[x][y]=='@')
{
return len;
}
else if(map[x][y]=='#')continue;
else if(map[x][y]=='.')
{
tmp.step=len+1;
if(visit[x-1][y])
{
tmp.x=x-1;
tmp.y=y;
qe.push(tmp);
visit[x-1][y]=0;
}
if(visit[x][y-1])
{
tmp.x=x;
tmp.y=y-1;
qe.push(tmp);
visit[x][y-1]=0;
}
if(visit[x][y+1])
{
tmp.x=x;
tmp.y=y+1;
qe.push(tmp);
visit[x][y+1]=0;
}
if(visit[x+1][y])
{
tmp.x=x+1;
tmp.y=y;
qe.push(tmp);
visit[x+1][y]=0;
}
}
}
return -1;
}
int dfs(int fx,int ty,int cnt)
{
int temp,minn=999999999;
vis[ty]=1;
if(cnt==q)
{
return dis[fx][ty];
}
for(int i=1;i<=q;i++)
{
if(!vis[i])
{
temp=dfs(ty,i,cnt+1);
// printf("%d\n",temp);
minn=min(temp,minn);
vis[i]=0;
}
}
return dis[fx][ty]+minn;
}
int main()
{
while(scanf("%d%d",&n,&m)&&(n||m))
{
memset(dis,-1,sizeof(dis));
bool flag=true;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
scanf(" %c",&map[i][j]);
}
scanf("%d",&q);
for(int i=1;i<=q;i++)
scanf("%d%d",&desx[i],&desy[i]);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
if(map[i][j]=='@')
{
desx[0]=i;
desy[0]=j;
}
}
map[desx[0]][desy[0]]='.';
for(int i=0;i<q;i++)
{
for(int j=0;j<=q-1-i;j++)
{
fmx=desx[i];
fmy=desy[i];
tox=desx[i+j+1];
toy=desy[i+j+1];
map[tox][toy]='@';
dis[i][j+i+1]=bfs();
map[tox][toy]='.';
}
}
for(int i=0;i<=q;i++)
{
for(int j=0;j<=q;j++)
{
tmp=max(dis[i][j],dis[j][i]);
dis[i][j]=dis[j][i]=tmp;
//printf("%d-%d: %d\n",i,j,dis[i][j]);
}
}
for(int i=0;i<=q;i++)
{
for(int j=0;j<=q;j++)
{
if(i!=j&&dis[i][j]==-1)
{
flag=false;
break;
}
}
}
if(!flag)
{
printf("-1\n");
continue;
}
ans=99999998;
//printf("**\n");
memset(vis,0,sizeof(vis));
for(int i=1;i<=q;i++)
{
ans=min(ans,dfs(0,i,1));
vis[i]=0;
}
//printf("***\n");
printf("%d\n",ans);
}
return 0;
}
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HDU 4771 Stealing Harry Potter's Precious （深搜+广搜）