CSU 1815: Enterprising Escape (BFS搜索)

1815: Enterprising Escape

    Time Limit: 20 Sec    Memory Limit: 1024 Mb    Submitted: 365    Solved: 117    

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Description

    The Enterprise is surrounded by Klingons! Find the escape route that has the quickest exit time, and print that time.

    Input is a rectangular grid; each grid square either has the Enterprise or some class of a Klingon warship. Associated with each class of Klingon warship is a time that it takes for the Enterprise to defeat that Klingon. To escape, the Enterprise must defeat each Klingon on some path to the perimeter. Squares are connected by their edges, not by corners (thus, four neighbors).

Input

    The first line will contain T, the number of cases; 2 ≤ T ≤ 100. Each case will start with line containing three numbers k, w, and h. The value for k is the number of different Klingon classes and will be between 1 and 25, inclusive. The value for w is the width of the grid and will be between 1 and 1000, inclusive. The value for h is the height of the grid and will be between 1 and 1000, inclusive.

    Following that will be k lines. Each will consist of a capital letter used to label the class of Klingon ships followed by the duration required to defeat that class of Klingon. The label will not be "E". The duration is in minutes and will be between 0 and 100,000, inclusive. Each label will be distinct.

    Following that will be h lines. Each will consist of w capital letters (with no spaces between them). There will be exactly one "E" across all h lines, denoting the location of the Enterprise; all other capital letters will be one of the k labels given above, denoting the class of Klingon warship in the square.

Output

Your output should be a single integer value indicating the time required for the Enterprise toescape.

Sample Input

26 3 3A 1B 2C 3D 4F 5G 6ABCFECDBG2 6 3A 100B 1000BBBBBBAAAAEBBBBBBB

Sample Output

2400

Hint

Source

2013 Pacific Northwest Region Programming Contest

简单的BFS搜索题，处理每个字母代表的数值，可以用数组将其转化成数字来处理，也可以直接用map来存储每个字母所代表的的信息

#include <iostream>#include <cstring>#include <stack>#include <cstdio>#include <cmath>#include <queue>#include <algorithm>#include <vector>#include <set>#include <map>const double eps=1e-8;const double PI=acos(-1.0);using namespace std;struct st{    int x;    int y;    int step;    friend bool operator <(st a,st b){    return a.step>b.step;}}node;char a[1005][1005];int v[1005][1005];int d[][2]={{1,0},{0,1},{0,-1},{-1,0}};int k,n,m,num; map<char,int> s;int bfs(int x,int y){    priority_queue<st>q;    node.x=x;    node.y=y;    node.step=0;    q.push(node);    v[x][y]=1;    while(!q.empty())    {        st tem=q.top();        //cout<<tem.x<<" "<<tem.y<<endl;        q.pop();        st tp;         if(tem.x>=m-1||tem.y>=n-1||tem.x<=0||tem.y<=0)            {                return tem.step;//cout<<"t:"<<e<<endl;            }        for(int i=0; i<4; i++)        {            tp.x=tem.x+d[i][0];            tp.y=tem.y+d[i][1];            if(v[tp.x][tp.y]) continue;            tp.step=tem.step+s[a[tp.x][tp.y]];            q.push(tp);            v[tp.x][tp.y]=1;            //cout<<tp<<endl;        }    }    return -1;}int main(){    int t;    scanf("%d",&t);    while(t--){            memset(v,0,sizeof(v));        char c;        scanf("%d%d%d",&k,&n,&m);        getchar();        for(int i=0;i<k;i++){            scanf("%c %d",&c,&num);            getchar();            s[c]=num;            }            int x,y;        for(int i=0;i<m;i++)            scanf("%s",a[i]);            for(int i=0;i<m;i++)                for(int j=0;j<n;j++)                if(a[i][j]=='E'){                    x=i;y=j;                }           printf("%d\n",bfs(x,y));    }    return 0;}