HDU 4308 Saving Princess claire_（简单BFS）

求出不使用P点时起点到终点的最短距离，求出起点到所有P点的最短距离，求出终点到所有P点的最短距离。

答案=min（ 不使用P点时起点到终点的最短距离， 起点到P的最短距离+终点到P的最短距离 ）

 

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <algorithm>

#include <queue>



using namespace std;



const int MAXN = 5100;

const int INF  = 1 << 30;



const int dx[] = { -1,  0, 0, 1 };

const int dy[] = {  0, -1, 1, 0 };



struct node

{

    int x, y;

    int cost;

    int id;

    bool operator<( const node &rhs ) const

    {

        return cost < rhs.cost;

    }

};



int R, C, cost;

char mat[MAXN][MAXN];

bool vis[MAXN][MAXN];

int  index[MAXN][MAXN];

int G[2][MAXN/10];

node P[MAXN/10], start, end;

int cntP;



void init( )

{

    cntP = 1;

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

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

        {

            if ( mat[i][j] == 'P' )

            {

                P[cntP].x = i;

                P[cntP].y = j;

                P[cntP].cost = 0;

                P[cntP].id = cntP;

                //printf( "cntP=%d [%d, %d]\n", cntP, i, j );

                ++cntP;

            }

            else if ( mat[i][j] == 'Y' )

            {

                start.x = i;

                start.y = j;

                start.cost = 0;

            }

            else if ( mat[i][j] == 'C' )

            {

                end.x = i;

                end.y = j;

                end.cost = 0;

            }

        }



    P[0] = start;

    P[0].id = 0;



    P[cntP] = end;

    P[cntP].id = cntP;

    //printf( "cntP = %d\n", cntP );

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

        index[ P[i].x ][ P[i].y ] = i;

    return;

}



bool check( int x, int y )

{

    if ( x >= 0 && x < R && y >= 0 && y < C ) return true;

    return false;

}



int BFS( int c, node st )

{

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

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

        vis[i][j] = false;



    queue<node> Q;



    st.cost = 0;

    vis[ st.x ][ st.y ] = true;

    Q.push(st);



    while ( !Q.empty() )

    {

        node cur = Q.front();

        Q.pop();



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

        {

            int xx = cur.x + dx[i];

            int yy = cur.y + dy[i];

            if ( check( xx, yy ) && mat[xx][yy] != '#' && !vis[xx][yy] )

            {

                node next;

                next.x = xx;

                next.y = yy;

                next.cost = cur.cost;

                next.id = index[xx][yy];

                if ( mat[xx][yy] != '*' )

                {

                    int id = index[xx][yy];

                    G[c][id] = next.cost;

                }

                else

                    next.cost += cost;



                vis[xx][yy] = true;

                Q.push(next);

            }

        }

    }

    return INF;

}



int getG()

{

    //printf("***cntP=%d\n", cntP);

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

    {

        G[0][i] = INF;

        G[1][i] = INF;

    }



    BFS( 0, P[0] );

    BFS( 1, P[cntP] );



    int ans = G[0][cntP];

    //printf("***ans=%d\n", ans );



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

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

        {

            if ( G[0][i] < INF && G[1][j] < INF )

                ans = min( ans, G[0][i] + G[1][j] );

            //printf( "ans = %d\n", ans );

        }



    return ans;

}



int main()

{

    while ( scanf( "%d%d%d", &R, &C, &cost ) == 3 )

    {

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

            scanf( "%s", mat[i] );



        init();

        int ans = getG();



        if ( ans < INF ) printf( "%d\n", ans );

        else puts("Damn teoy!");

    }

    return 0;

}