acmdream1191 bfs+優先佇列

life4711發表於2014-09-23
ACM佇列

http://115.28.76.232/problem?pid=1191

Problem Description

You are the prince of Dragon Kingdom and your kingdom is in danger of running out of power. You must find power to save your kingdom and its people. An old legend states that power comes from a place known as Dragon Maze. Dragon Maze appears randomly out of nowhere without notice and suddenly disappears without warning. You know where Dragon Maze is now, so it is important you retrieve some power before it disappears.

Dragon Maze is a rectangular maze, an N×M grid of cells. The top left corner cell of the maze is (0, 0) and the bottom right corner is(N-1, M-1). Each cell making up the maze can be either a dangerous place which you never escape after entering, or a safe place that contains a certain amount of power. The power in a safe cell is automatically gathered once you enter that cell, and can only be gathered once. Starting from a cell, you can walk up/down/left/right to adjacent cells with a single step.

Now you know where the entrance and exit cells are, that they are different, and that they are both safe cells. In order to get out of Dragon Maze before it disappears, you must walk from the entrance cell to the exit cell taking as few steps as possible.

If there are multiple choices for the path you could take, you must choose the one on which you collect as much power as possible in order to save your kingdom.

Input

The first line of the input gives the number of test cases, T(1 ≤ T ≤ 30). T test cases follow.

Each test case starts with a line containing two integers N and M(1 ≤ N, M ≤ 100), which give the size of Dragon Maze as described above.

The second line of each test case contains four integers enx, eny, exx, exy(0 ≤ enx, exx < N, 0 ≤ eny, exy < M), describing the position of entrance cell (enx, eny) and exit cell (exx, exy).

Then N lines follow and each line has M numbers, separated by spaces, describing the N×M cells of Dragon Maze from top to bottom.

Each number for a cell is either -1, which indicates a cell is dangerous, or a positive integer, which indicates a safe cell containing a certain amount of power.

Output

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1).

If it's possible for you to walk from the entrance to the exit, y should be the maximum total amount of power you can collect by taking the fewest steps possible.

If you cannot walk from the entrance to the exit, y should be the string "Mission Impossible." (quotes for clarity).

Sample Input

2
2 3
0 2 1 0
2 -1 5
3 -1 6
4 4
0 2 3 2
-1 1 1 2
1 1 1 1
2 -1 -1 1
1 1 1 1

Sample Output

Case #1: Mission Impossible.
Case #2: 7
解題思路：

              剛開始的時候沒想到用優先佇列，採用了普通的佇列直接進行了bfs後來才想到自己忘記了寫回溯，這樣一來先搜到的路如果不是最優解的話一旦標記就把後面的本來可行的路給封死了，無疑就得不到正解了，因此我們在搜尋的時候一定要保證先搜到的肯定是最優解，想到這裡就很容易想到優先佇列了，那麼題目就迎刃而解了

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;

struct note
{
    int x,y,num,sum;
    note() {}
    note(int _x,int _y,int _num,int _sum):x(_x),y(_y),num(_num),sum(_sum) {}
    bool operator <(const note a)const
    {
        return (a.num<num||(a.num==num&&sum<a.sum));
    }
};

int dx[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
int T,n,m,x1,x2,y1,y2,a[150][150];
bool flag[150][150];
priority_queue<note>q;

bool judge(int x,int y)
{
    if(x<n&&x>=0&&y<m&&y>=0&&a[x][y]!=-1)
        return true;
    return false;
}

bool bfs(note p)
{
    memset(flag,0,sizeof(flag));
    while(!q.empty())
        q.pop();
    q.push(p);
    while(!q.empty())
    {
        note p=q.top();
        int x=p.x;
        int y=p.y;
        int num=p.num;
        int sum=p.sum;
        if(x==x2&&y==y2)
        {
            printf("%d\n",sum);
            return true;
        }
        q.pop();
        for(int i=0; i<4; i++)
        {
            int xx=x+dx[i][0];
            int yy=y+dx[i][1];
            if(judge(xx,yy)&&flag[xx][yy]==0)
            {
                q.push(note(xx,yy,num+1,sum+a[xx][yy]));
                flag[xx][yy]=1;
            }
        }
    }
    return false;
}
int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                scanf("%d",&a[i][j]);
        printf("Case #%d: ",++tt);
        if(!bfs(note(x1,y1,0,a[x1][y1])))
            printf("Mission Impossible.\n");
    }
    return 0;
}


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