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  • 搜索

    见破题:

    思路十分简单,至于代码嘛...反正我是改了一上午...

    这里就想说一个搜索的通用方法,可以用f[i][j]表示某个状态,然后如果搜索到相同的坐标,但价值没有之前存的好,就直接退出...

    至于此题就是(i,j)点的最大价值...

    #include<bits/stdc++.h>
    #define max(a,b) (((a)>(b)?(a):(b)))
    using namespace std;
    int h,v,w,qx,qy,zx,zy,dx[5]={1,2,3,4},d,f[50][50][5],ans;//f[i][j]表示到(i,j)点的最大速度. 
    char ch[50][50];//1-左,2-右 
    inline int read()
    {
        int x=0,ff=1;
        char ch=getchar();
        while(!isdigit(ch)){if(ch=='-') ff=-1;ch=getchar();}
        while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
        return x*ff; 
    }
    inline void dfs(int x,int y,int d,int v,int m)
    {
        //printf("x=%d y=%d d=%d v=%d
    ",x,y,d,v);
        if(v<=f[x][y][d]) return;f[x][y][d]=v;
        if(ch[x][y]=='F') return;
        if(d==1)
        {
            int pd=0;
            if(x-1>=1&&ch[x-1][y]!='.') pd=1,dfs(x-1,y,1,v+1,0);   
            if(m!=1&&y-1>=1&&ch[x][y-1]!='.') pd=1,dfs(x,y,4,max(0,v-35),1);
             if(m!=2&&y+1<=w&&ch[x][y+1]!='.') pd=1,dfs(x,y,2,max(0,v-40),2);
            if(!pd&&x+1<=h&&ch[x+1][y]!='.') dfs(x,y,3,0,0);
        }
        else if(d==2)
        {
            int pd=0;
            if(y+1<=w&&ch[x][y+1]!='.') pd=1,dfs(x,y+1,2,v+1,0);
            if(m!=1&&x-1>=1&&ch[x-1][y]!='.') pd=1,dfs(x,y,1,max(0,v-35),1);
            if(m!=2&&x+1<=h&&ch[x+1][y]!='.') pd=1,dfs(x,y,3,max(0,v-40),2);
            if(!pd&&y-1>=1&&ch[x][y-1]!='.') dfs(x,y,4,0,0);
        }
        else if(d==3)
        {
            int pd=0;
            if(x+1<=h&&ch[x+1][y]!='.')pd=1,dfs(x+1,y,3,v+1,0);
            if(m!=1&&y+1<=w&&ch[x][y+1]!='.')pd=1,dfs(x,y,2,max(0,v-35),1);
             if(m!=2&&y-1>=1&&ch[x][y-1]!='.')pd=1,dfs(x,y,4,max(0,v-40),2);
            if(!pd&&x-1>=1&&ch[x-1][y]!='.') dfs(x,y,1,0,0);
        }
        else if(d==4)
        {
            int pd=0;
            if(y-1>=1&&ch[x][y-1]!='.')pd=1,dfs(x,y-1,4,v+1,0);
            if(m!=1&&x+1<=h&&ch[x+1][y]!='.')pd=1,dfs(x,y,3,max(0,v-35),1);   
            if(m!=2&&x-1>=1&&ch[x-1][y]!='.')pd=1,dfs(x,y,1,max(0,v-40),2);
            if(!pd&&y+1<=w&&ch[x][y+1]!='.') dfs(x,y,2,0,0);
        }
    }
    int main()
    {
        freopen("1.in","r",stdin);
        h=read();w=read();v=read();
        for(register int i=1;i<=h;++i)
            for(register int j=1;j<=w;++j)  
            {
                cin>>ch[i][j];
                if(ch[i][j]=='F') zx=i,zy=j;
                else if(ch[i][j]!='.'&&ch[i][j]!='#')
                {
                    qx=i;qy=j;
                    if(ch[i][j]=='N') d=1;
                    else if(ch[i][j]=='E') d=2;
                    else if(ch[i][j]=='S') d=3;
                    else if(ch[i][j]=='W') d=4;
                }     
            }    
        for(register int i=1;i<=h;++i)
            for(register int j=1;j<=w;++j) 
                for(register int k=1;k<=4;k++) f[i][j][k]=-1e9;     
        dfs(qx,qy,d,v,0);    
        for(register int i=1;i<=4;++i) ans=max(ans,f[zx][zy][i]);
        printf("%d
    ",ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/gcfer/p/11607241.html
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