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]
", 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
", 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
", 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
", 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
", 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
", ans );
        else puts("Damn teoy!");
    }
    return 0;
}