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.

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).

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

Case #1: Mission Impossible.

Case #2: 7

意解: 一个简单的BFS加优先队列,dfs也能够做,毕竟数据非常小.

AC代码:

            

/*

*  this code is made by eagle

*  Problem: 1191

*  Verdict: Accepted

*  Submission Date: 2014-09-19 01:46:08

*  Time: 72MS

*  Memory: 1736KB

*/

#include <iostream>

#include <cstring>

#include <cstdio>

#include <queue>

using namespace std;

int N,M,ux,uy,nx,ny;

int map[110][110];

int d[][2] = {-1,0,1,0,0,1,0,-1};

struct Node

{

    int x,y,num,t;

    //Node (int x, int y, int num) : x(x),y(y),num(num) {}

    bool operator < (const Node &cnt) const

    {

        return cnt.t < t || (cnt.num > num && cnt.t == t);

    }

};

bool query()

{

    Node a;

    priority_queue<Node>ms;

    a.x = ux;

    a.y = uy;

    a.num = map[ux][uy];

    a.t = 0;

    ms.push(a);

    while(!ms.empty())

    {

        a = ms.top();

        ms.pop();

//        cout<<a.x<<":"<<a.y<<endl;

        if(a.x == nx && a.y == ny)

        {

            printf("%d\n",a.num);

            return true;

        }

        for(int i = 0; i < 4; i++)

        {

            Node b;

            b.x = a.x + d[i][0];

            b.y = a.y + d[i][1];

            if(b.x < 0 || b.y < 0 || b.x >= N || b.y >= M || map[b.x][b.y] == -1) continue;

            b.num = a.num + map[b.x][b.y];

            b.t = a.t + 1;

            ms.push(b);

            map[b.x][b.y] = -1;

        }

    }

    return false;

}

int main()

{

    //freopen("in.txt","r",stdin);

    int T,cnt = 0;

    scanf("%d",&T);

    while(T--)

    {

        printf("Case #%d: ",++cnt);

        scanf("%d %d",&N,&M);

        scanf("%d %d %d %d",&ux,&uy,&nx,&ny);

        for(int i = 0; i < N; i++)

            for(int j = 0; j < M; j++)

                scanf("%d",&map[i][j]);

        if(!query())

            puts("Mission Impossible.");

    }

    return 0;

}

5

A d 6

B d 3

C e 9

d Z 8

e Z 3

B 11

意解: 这是一个简单的floyd最短路算法,仅仅须要给每一个村庄表上号,以A-z标为1到64,然后就是一个模板题了.

AC代码:

/*

* this code is made by eagle

* Problem: 1209

* Verdict: Accepted

* Submission Date: 2014-09-20 10:58:32

* Time: 12MS

* Memory: 1716KB

*/

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

const int INF = 1e8;

int map[100][100];

void unit()

{

    for(int i = 1; i <= 100; i++)

        for(int j = 1; j <= 100; j++)

            if(i != j) map[i][j] = INF;

            else map[i][j] = 0;

}

int main()

{

    //freopen("in","r",stdin);

    int T;

    cin>>T;

    getchar();

    unit();

    while(T--)

    {

        char c1,c2;

        int x,a,b;

        scanf("%c %c %d",&c1,&c2,&x);

        getchar();

        if(c1 > c2) swap(c1,c2);

        if(c1 > 'Z') a = c1 - 'a' + 27;

        else a = c1 - 'A' + 1;

        if(c2 > 'Z') b = c2 - 'a' + 27;

        else b = c2 - 'A' + 1;

        if(x < map[a][b]) map[a][b] = map[b][a] = x;

    }

    for(int k = 1; k <= 70; k++)

        for(int j = 1; j <= 70; j++)

            for(int i = 1; i <= 70; i++)

                if(map[j][k] == INF || map[k][i] == INF) continue;

                else map[j][i] = min(map[j][i] , map[j][k] + map[k][i]);

    int ans = INF;

    char t;

    for(int i = 1; i < 26; i++)

        if(map[i][26] < ans)

        {

            t = (char)(i + 'A' - 1);

            ans = map[i][26];

        }

    printf("%c %d\n",t,ans);

    return 0;

}