【spfa】【動態規劃】zoj3847 Collect Chars

轉載自：http://blog.csdn.net/madaidao/article/details/42616743

 

 

Collect Chars

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Bob was playing MC and was punished by Alice. Bob was trapped in a maze and there were some characters on some specific cells. When he walks on a specific point, he will pick up the character automatically (he cannot refuse to do that) and push back the character in his bag (You can treat it as adding a character to the end of a string and initially, the string is empty.). Alice has set several goal strings and when Bob's bag contains any one of the strings, Bob is enabled to return to the true world.

Bob can walk to the adjacent cell (up, down, left, right) at one time, and it costs no time to pick up the character and the character won't disappear, which means if Bob want's he can get endless characters once he walked on the certain cell but he must pick up at least one character.

Bob wants to leave the maze as soon as possible, and he asks you for the shortest time that he can go out of that maze.

Input

The first line is an integer n, indicating the number of testcases.

In each case, the first line is two integers L and C (0 < L, C ≤ 20). After this, L lines follow, and each has Ccharacters. For each character, '.' stands for an open space, '#' stands for a wall which cannot be crossed, '@' stands for the beginning point of the maze. All capital letters 'A' to 'Z' stand for the characters which can be picked up.

After that, there is one integer W, which means the number of goal strings (0 < W ≤ 200). And W lines following, each line has a string, containing letters 'A' to 'Z' only, which length is less than 100.

Output

For each case, you should output an integer. The answer is the minimum time that Bob must use to get out of the maze. If he can't do that, you should output -1.

Sample Input

2
5 5
A.DB#
@#.##
.#..C
.#...
.....
3
AB
AC
BC
5 5
A.DB#
@#.##
.#..C
.#...
.....
2
AADBBBDCC
AC

Sample Output

11
9

 

題意：BOb在一個L*C的格子迷宮中，有些特別的格子裏面有一個大寫字母，每次Bob走到這個格子，至少要添加一個該字母到包裏，至關於添加一個字母到字符串的末尾，初始的時候字符串爲空。能夠添加無數個（字母拿了不會消失，拿字母不花時間）。Alice有一些字符串，如有一個字符串爲Bob包裏的串的子串，Bob就能夠走出迷宮。求Bpb走出迷宮的最小時間。

題解：枚舉Alice的串，求出靠每一個Alice的串走出迷宮的最小時間。對於每一個串s，咱們用dp[i][j][k] 表示當前在點(i,j)，揹包裏的串的長度爲j的後綴和s的長度爲k的前綴相同，走到該狀態的最短期。用spfa的方法轉移便可。代碼以下：

 

#include<stdio.h>  
#include<iostream>  
#include<algorithm>  
#include<string.h>  
#include<string>  
#include<queue>  
#include<stack>  
#include<map>  
#include<set>  
#include<stdlib.h>  
#include<vector>  
#define inff 0x3fffffff  
#define nn 110000  
#define mod 1000000007  
typedef long long LL;  
const LL inf64=inff*(LL)inff;  
using namespace std;  
int c,l;  
int w;  
int dir[4][2]={1,0,-1,0,0,1,0,-1};  
char tu[50][50];  
LL dp[50][50][1100];  
bool inque[50][50][1100];  
struct node  
{  
    int x,y,zt;  
    node(){}  
    node(int xx,int yy,int zzt)  
    {  
        x=xx,y=yy,zt=zzt;  
    }  
}sta,tem;  
queue<node>que;  
void gengxin(int x,int y,int zt,int val)  
{  
    if(dp[sta.x][sta.y][sta.zt]+val<dp[x][y][zt])  
    {  
        dp[x][y][zt]=dp[sta.x][sta.y][sta.zt]+val;  
        if(!inque[x][y][zt])  
        {  
            inque[x][y][zt]=true;  
            que.push(node(x,y,zt));  
        }  
    }  
}  
LL solve(string s)  
{  
    int ls=s.size();  
    int i,j,k;  
    for(i=0;i<c;i++)  
    {  
        for(j=0;j<l;j++)  
        {  
            for(k=0;k<=ls;k++)  
            {  
                inque[i][j][k]=false;  
                dp[i][j][k]=inf64;  
            }  
        }  
    }  
    while(que.size())  
        que.pop();  
    for(i=0;i<c;i++)  
    {  
        for(j=0;j<l;j++)  
        {  
            if(tu[i][j]=='@')  
            {  
                dp[i][j][0]=0;  
                inque[i][j][0]=true;  
                que.push(node(i,j,0));  
            }  
        }  
    }  
    int x,y;  
    LL re=inf64;  
    while(que.size())  
    {  
        sta=que.front();  
        que.pop();  
        inque[sta.x][sta.y][sta.zt]=false;  
        if(sta.zt==ls)  
        {  
            re=min(re,dp[sta.x][sta.y][sta.zt]);  
            continue;  
        }  
        if(tu[sta.x][sta.y]==s[sta.zt])  
        {  
            gengxin(sta.x,sta.y,sta.zt+1,0);  
        }  
        for(i=0;i<4;i++)  
        {  
            x=sta.x+dir[i][0];  
            y=sta.y+dir[i][1];  
            if(x>=0&&x<c&&y>=0&&y<l)  
            {  
                if(tu[x][y]==s[sta.zt])  
                {  
                    gengxin(x,y,sta.zt+1,1);  
                }  
                if(tu[x][y]=='@'||tu[x][y]=='.')  
                {  
                    gengxin(x,y,sta.zt,1);  
                }  
                if(tu[x][y]>='A'&&tu[x][y]<='Z'&&tu[x][y]!=s[sta.zt])  
                {  
                    gengxin(x,y,0,1);  
                }  
            }  
        }  
    }  
    return re;  
}  
char sr[1100];  
int main()  
{  
    int t,i;  
    string s;  
    scanf("%d",&t);  
    while(t--)  
    {  
        scanf("%d%d",&c,&l);  
        for(i=0;i<c;i++)  
        {  
            scanf("%s",tu[i]);  
        }  
        scanf("%d",&w);  
        getchar();  
        LL ans=inf64;  
        for(i=1;i<=w;i++)  
        {  
            gets(sr);  
            s=sr;  
            ans=min(ans,solve(s));  
        }  
        if(ans==inf64)  
            printf("%d\n",-1);  
        else  
            printf("%lld\n",ans);  
    }  
    return 0;  
}