HDU 4308-saving princess Claire _

 

A huge string of text is actually useless.

Typical maze questions. The prince rescued the princess.

Y is King y Sub, C is the princess, p is the transfer array (can be instantly transmitted to any one for free), # Is the wall, * It is a toll station (it will cost money every time it passes through ).

Q: Can I save the princess? What is the minimum cost for saving the money?

No special P. priority queue is used directly...

Although the use of priority queue is a waste, but fortunately the code is the shortest. Fast.

 

#include<iostream>#include<cstdio>#include<queue>#include<algorithm>#include<cstring>#include<cmath>using namespace std;struct point{    int x,y;}p[1000];struct path{    int x,y,cost;    path(){};    path(int a,int b,int c){        x=a;        y=b;        cost=c;    }    bool friend operator<(path p,path q){        return p.cost>q.cost;    }};priority_queue<path>que;int n,m,cost;int fx[]={1,-1,0,0};int fy[]={0,0,1,-1};string map[5004];int yx,yy;int cnt;int bfs(){    int k,i,j;    path no,nx;    while(!que.empty()) que.pop();    que.push(path(yx,yy,0));    map[yx][yy]='#';    while(!que.empty()){        no=que.top();        que.pop();        for(k=0;k<4;k++){            nx.x=no.x+fx[k];            nx.y=no.y+fy[k];            if(nx.x<0 || nx.x>=n || nx.y<0 || nx.y>=m) continue;            if(map[nx.x][nx.y]=='#') continue;            if(map[nx.x][nx.y]=='P'){                for(i=0;i<cnt;i++){                    que.push(path(p[i].x,p[i].y,no.cost));                    map[p[i].x][p[i].y]='#';                }            }            else{                if(map[nx.x][nx.y]=='C')                    return no.cost;                else{                    map[nx.x][nx.y]='#';                    que.push(path(nx.x,nx.y,no.cost+cost));                }            }        }    }    return -1;}int main(){    int i,j,res;    while(cin>>n>>m>>cost){        cnt=0;        for(i=0;i<n;i++){            cin>>map[i];            for(j=0;j<m;j++)                if(map[i][j]=='Y')                    yx=i,yy=j;                else if(map[i][j]=='P'){                    p[cnt].x=i;                    p[cnt].y=j;                    cnt++;                }        }        res=bfs();        if(res==-1)            cout<<"Damn teoy!"<<endl;        else            cout<<res<<endl;    }    return 0;}