快速切题 hdu2416 Treasure of the Chimp Island 搜索 解题报告
Treasure of the Chimp Island
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 313 Accepted Submission(s): 151
The hardness of the stone blocks is an integer between 1 and 9, showing the number of days required to destroy the block. We neglect the time required to travel inside the corridors. Using a dynamite, Bob can destroy a block almost immediately, so we can ignore the time required for it too. The problem is to find the minimum time at which Bob can reach the treasure. He may choose any gate he wants to enter ZM2.
*.1....4..$...*
*..***..2.....*
*..2..*****..2*
*..3..******37A
*****9..56....*
*.....******..*
***CA**********
*****
*$3**
*.2**
***#*
--
IMPOSSIBLE
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=111;
char maz[maxn][maxn];
int n,m;
int vis[maxn][maxn][27];
const int inf=0x7ffffff;
class node{
public:
int x,y,t,p;
node(){x=y=t=p=0;}
node (int tx,int ty,int tt,int tp):x(tx),y(ty),t(tt),p(tp){}
bool operator <(const node & n2)const {
return t>n2.t;
}
};
int isdoor(int x,int y){
if(maz[x][y]=='#')return 0;
if(maz[x][y]>='A'&&maz[x][y]<='Z')return maz[x][y]-'A'+1;
return -1;
}
int isstone(int x,int y){
if(maz[x][y]>='0'&&maz[x][y]<='9')return maz[x][y]-'0';
return -1;
}
bool judge(int x,int y){
if(x>=0&&x<n&&y>=0&&y<m)return true;
return false;
}
void printmaz(){
printf("maz %d %d\n",n,m);
for(int i=0;i<n;i++)printf("%s\n",maz[i]);
}
priority_queue <node >que;
const int dx[4]={1,-1,0,0},dy[4]={0,0,1,-1};
int bfs(){
while(!que.empty())que.pop();
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(isdoor(i,j)!=-1){
que.push(node(i,j,0,isdoor(i,j)));
maz[i][j]='*';
}
}
}
while(!que.empty()){
node tp=que.top();que.pop();
//printf("pack x%d y%d t%d p%d\n",tp.x,tp.y,tp.t,tp.p);
for(int i=0;i<4;i++){
int tx=tp.x+dx[i],ty=tp.y+dy[i];
if(judge(tx,ty)){
if(maz[tx][ty]=='$')return tp.t;
else if(maz[tx][ty]=='.'){
if(vis[tx][ty][tp.p]==-1||vis[tx][ty][tp.p]>tp.t){
vis[tx][ty][tp.p]=tp.t;
que.push(node(tx,ty,tp.t,tp.p));
}
}
else if(isstone(tx,ty)!=-1){
if(vis[tx][ty][tp.p]==-1||vis[tx][ty][tp.p]>tp.t+isstone(tx,ty)){
vis[tx][ty][tp.p]=tp.t+isstone(tx,ty);
que.push(node(tx,ty,tp.t+isstone(tx,ty),tp.p));
}
if(tp.p>0&&(vis[tx][ty][tp.p-1]==-1||vis[tx][ty][tp.p-1]>tp.t)){
vis[tx][ty][tp.p-1]=tp.t;
que.push(node(tx,ty,tp.t,tp.p-1));
}
}
}
}
}
return inf;
}
int main(){
while(1){
for(n=0;(gets(maz[n]))&&strcmp(maz[n],"--")!=0&&strlen(maz[n])!=0;n++){}
if(strlen(maz[n])!=0)break;//in the end
m=strlen(maz[0]);
memset(vis,-1,sizeof(vis));
int ans;
ans=bfs();
if(ans!=inf)printf("%d\n",ans);
else puts("IMPOSSIBLE");
}
return 0;
}
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