ACM-icpc---bfs求最短路

You are working on the team assisting with programming for the Mars rover. To conserve energy, the rover needs to find optimal paths across the rugged terrain to get from its starting location to its final location. The following is the first approximation for the problem.


N * N square matrices contain the expenses for traversing each individual cell. For each of them, your task is to find the minimum-cost traversal from the top left cell [0][0] to the bottom right cell [N-1][N-1]. Legal moves are up, down, left, and right; that is, either the row index changes by one or the column index changes by one, but not both.

 

Input

Each problem is specified by a single integer between 2 and 125 giving the number of rows and columns in theN * N square matrix. The file is terminated by the caseN = 0.

Following the specification of N you will find N lines, each containingN numbers. These numbers will be given as single digits, zero through nine, separated by single blanks.

 

Output

Each problem set will be numbered (beginning at one) and will generate a single line giving the problem set and the expense of the minimum-cost path from the top left to the bottom right corner, exactly as shown in the sample output (with only a single space after "Problem" and after the colon).

 

Sample Input

题意：给出一个n阶的矩阵，求从a[0][0]到a[n-1][n-1]的最短距离；

35 5 43 9 13 2 753 7 2 0 12 8 0 9 11 2 1 8 19 8 9 2 03 6 5 1 579 0 5 1 1 5 34 1 2 1 6 5 30 7 6 1 6 8 51 1 7 8 3 2 39 4 0 7 6 4 15 8 3 2 4 8 37 4 8 4 8 3 40

 

Sample Output

Problem 1: 20Problem 2: 19Problem 3: 36
#include<cstdio>#include<queue>#include<cstring>#include<string>#include<algorithm>using namespace std;int dx[]= {-1,0,1,0};int dy[]= {0,1,0,-1};int a[150][150],vis[150][150];int n,ans;struct A{    int x,y;    int dis;    friend bool operator <(A a,A b)    {        return a.dis>b.dis;    }};void bfs(){    priority_queue<A> Q;    A f;    f.x=0;    f.y=0;    f.dis=a[0][0];    vis[0][0]=1;    Q.push(f);    while(!Q.empty())    {        f=Q.top();        Q.pop();        if(f.x==n-1&&f.y==n-1)        {            ans=f.dis;            return;        }        for(int i=0;i<4; i++)        {            A p=f;            p.x+=dx[i];            p.y+=dy[i];            p.dis+=a[p.x][p.y];            if(p.x>=0&&p.x<n&&p.y>=0&&p.y<n&&!vis[p.x][p.y])            {                vis[p.x][p.y]=1;                Q.push(p);            }        }    }}int main(){    int cas=1;    while(~scanf("%d",&n),n)    {        memset(vis,0,sizeof(vis));        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                scanf("%d",&a[i][j]);        ans=0;        bfs();        printf("Problem %d: %d\n",cas++,ans);    }    return 0;}