HDU 1044(BFS+DFS)

题目意思  求解是否能够到达出口 如果能 求解到达时的最大携带价值

 

首先使用广搜搜出包括起点和终点在内 所有特殊点之间的最短距离 建立一个隐式图

 

然后使用DFS枚举各个组合 然后求出最大值 注意DFS的强剪

 

以下是代码:

 

 

#include <iostream>
#include <queue>
#include <cmath>
using namespace std;

long Price[20];
long Step[20][20];
char Map[60][60];
long W,H,L,M;

typedef struct
{
    long mi,mj;
    long step;
    long from;
}Node;

queue<Node> q;

bool hash[60][60][20];
long dx[]={0,1,-1,0};
long dy[]={1,0,0,-1};


long b;

Node Pt[20];

inline void BFS()
{
    while (!q.empty())
    {
        q.pop();
    }
    memset(hash,0,sizeof(hash));

    q.push(Pt[0]);
    hash[Pt[0].mi][Pt[0].mj][Pt[0].from]=true;


    while (!q.empty())
    {
        Node t=q.front();
        q.pop();

        long j;
        Node n;

        for (j=0;j<4;++j)
        {
            n.mi=t.mi+dx[j];
            n.mj=t.mj+dy[j];
            n.step=t.step+1;
            n.from=t.from;
            if (n.mi>=0&&n.mi<H&&n.mj>=0&&n.mj<W)
            {
                if (n.step<=L&&!hash[n.mi][n.mj][n.from]&&Map[n.mi][n.mj]!='*')
                {
                    hash[n.mi][n.mj][n.from]=true;
                    q.push(n);

                    if (Map[n.mi][n.mj]>='A'&&Map[n.mi][n.mj]<='J')
                    {
                        Step[n.from][Map[n.mi][n.mj]-'A'+2]=Step[Map[n.mi][n.mj]-'A'+2][n.from]=n.step;
                        n.from=Map[n.mi][n.mj]-'A'+2;
                        n.step=0;
                        hash[n.mi][n.mj][n.from]=true;
                        q.push(n);    
                    }                
                    else if (Map[n.mi][n.mj]=='@')
                    {
                        Step[n.from][0]=Step[0][n.from]=n.step;
                        n.from=0;
                        n.step=0;
                        hash[n.mi][n.mj][n.from]=true;
                        q.push(n);    
                    }
                    else if (Map[n.mi][n.mj]=='<')
                    {
                        Step[n.from][1]=Step[1][n.from]=n.step;
                        n.from=1;
                        n.step=0;
                        hash[n.mi][n.mj][n.from]=true;
                        q.push(n);    
                    }
                }
            }

        }
    }
}

bool vist[20];
long lmax;
long all;

inline void dfs(long pos,long now,long step)
{
    if (lmax==all||step>L)
    {
        return;
    }
    if (Step[pos][1]!=-1&&step+Step[pos][1]<=L&&now>lmax)
    {
        lmax=now;
    }
    long i;
    for (i=2;i<M+2;++i)
    {
        if (!vist[i]&&Step[pos][i]!=-1)
        {
            vist[i]=true;
            dfs(i,now+Price[i],step+Step[pos][i]);
            vist[i]=false;
        }
    }
}


int main()
{
    long T;
    scanf("%ld",&T);
    b=1;
    while (T--)
    {
        scanf("%ld%ld%ld%ld",&W,&H,&L,&M);

        long i,j;

        all=0;
        for (i=2;i<M+2;++i)
        {
            scanf("%ld",&Price[i]);
            all+=Price[i];
        }
        Price[0]=Price[1]=0;

        gets(Map[0]);
        for (i=0;i<H;++i)
        {
            gets(Map[i]);
            for (j=0;j<W;++j)
            {
                if (Map[i][j]=='@')
                {
                    Pt[0].mi=i;
                    Pt[0].mj=j;
                    Pt[0].step=0;
                    Pt[0].from=0;
                }
                else
                if (Map[i][j]>='A'&&Map[i][j]<='J')
                {
                    Pt[Map[i][j]-'A'+2].mi=i;
                    Pt[Map[i][j]-'A'+2].mj=j;
                    Pt[Map[i][j]-'A'+2].step=0;
                    Pt[Map[i][j]-'A'+2].from=Map[i][j]-'A'+2;
                }
                else
                if (Map[i][j]=='<')
                {
                    Pt[1].mi=i;
                    Pt[1].mj=j;
                    Pt[1].step=0;
                    Pt[1].from=1;
                }
            }
        }

        memset(Step,-1,sizeof(Step));

        BFS();

        if (b!=1)
        {
            puts("");
        }
        printf("Case %ld:\n",b++);

        if (Step[0][1]==-1||Step[0][1]>L)
        {
            goto l1;
        }

        memset(vist,0,sizeof(vist));
        lmax=-1;
        vist[0]=true;
        dfs(0,0,0);



        if (lmax!=-1)
        {
            printf("The best score is %ld.\n",lmax);
        }
        else
        {
l1:
            puts("Impossible");
        }
                
    }
    return 0;
}