Sicily 4421. Adjacent Rooms-优快云博客

本文链接：https://blog.youkuaiyun.com/Ederick/article/details/7250065

本文分享了一名编程新手通过学习并应用Prim算法解决实际问题的过程，从不会到掌握，使用优先队列优化算法实现最小生成树的构建，通过实例演示了算法的具体应用。

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新手赛正赛热身题，当时不会做，连最少生成树是什么都不知道。后来上完课才知道这个算法，时隔多天后于寒假时做出来的……
总之就是，Prim最小生成树，保险起见还用了优先队列，其余的无需多讲。

Run Time: 0sec

Run Memory: 304KB

Code length: 1672Bytes

SubmitTime: 2012-01-26 17:13:23

// Problem#: 4421
// Submission#: 1200060
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <cstdio>
#include <queue>
using namespace std;

int dirt[ 4 ][ 2 ] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };

struct Pos {
    int x, y;
    int value;
    Pos() { x = 0; y = 0; value = 0; }
    friend bool operator < ( const Pos& p1, const Pos& p2 ) { return p1.value > p2.value; }
};

int square( int n ) { return n * n; }

int main()
{
    int i, j;
    int T, R, C;
    int rooms[ 51 ][ 51 ];
    bool visited[ 51 ][ 51 ];
    Pos add, temp;
    int count, sum;

    scanf( "%d", &T );
    while ( T-- ) {
        scanf( "%d%d", &R, &C );
        for ( i = 1; i <= R; i++ ) {
            for ( j = 1; j <= C; j++ ) {
                visited[ i ][ j ] = false;
                scanf( "%d", &rooms[ i ][ j ] );
                if ( rooms[ i ][ j ] == 0 ) {
                    add.x = i;
                    add.y = j;
                }
            }
        }

        priority_queue<Pos> q;
        sum = 0;
        for ( count = 1; count < R * C; count++ ) {
            visited[ add.x ][ add.y ] = true;
            for ( i = 0; i < 4; i++ ) {
                temp.x = add.x + dirt[ i ][ 0 ];
                temp.y = add.y + dirt[ i ][ 1 ];
                if ( temp.x >= 1 && temp.x <= R && temp.y >= 1 && temp.y <= C ) {
                    temp.value = square( rooms[ add.x ][ add.y ] - rooms[ temp.x ][ temp.y ] );
                    q.push( temp );
                }
            }

            add = q.top();
            while ( visited[ add.x ][ add.y ] ) {
                q.pop();
                add = q.top();
            }
            sum += add.value;
            q.pop();
        }
        printf( "%d\n", sum );
    }

    return 0;

}