SRM 583 Div II Level Three：GameOnABoard，Dijkstra最短路径算法

题目来源：http://community.topcoder.com/stat?c=problem_statement&pm=12556

用Dijkstra实现，之前用Floyd算法写了一个，结果在2s内算不出结果来。

参考了别人算法，学到了set容器的一个用法，用set省去了查找Dijkstra算法中选择最短路径的那一步，set中的第一个元素就是最小值，用priority queue应该也可以。

 

#include <iostream>

#include <string>

#include <vector>

#include <set>

#include <algorithm>



using namespace std;



#define MAX_SIZE 40

#define INFINITY 100000



#define entry(x, y) make_pair(dd[x][y], make_pair(x, y))



int dijkstra(int x, int y);

vector <string> bcost;



int dd[MAX_SIZE][MAX_SIZE];

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

int rows, cols;

class GameOnABoard

{

public:

	int optimalChoice(vector <string> cost);

};



int GameOnABoard::optimalChoice(vector<string> cost)

{

	int rows, cols, minL;

	rows = cost.size();

	cols = cost[0].size();

	minL = INFINITY;

	bcost = cost;

	for (int i = 0; i < rows; i++) {

		for (int j = 0; j < cols; j++) {

			minL = min(minL, dijkstra(i, j));

		}

	}



	return minL;

}



int dijkstra(int x, int y)

{

	int i, j, dir, maxC;

	int cus_x, cus_y, next_x, next_y;

	set < pair<int, pair<int, int>> > s;

	rows = bcost.size();

	cols = bcost[0].size();



	for (i = 0; i < rows; i++) {

		for (j = 0; j < cols; j++) {

			dd[i][j] = INFINITY;

		}

	}

	dd[x][y] = bcost[x][y] - '0';

	s.insert(entry(x, y));



	while (!s.empty()) {

		cus_x = s.begin()->second.first;

		cus_y = s.begin()->second.second;

		s.erase(s.begin());



		for (dir = 0; dir < 4; dir++) {

			next_x = cus_x + dx[dir];

			next_y = cus_y + dy[dir];

			if (next_x < 0 || next_x >= rows || next_y < 0 || next_y >= cols) {

				continue;

			}



			if (dd[next_x][next_y] <= dd[cus_x][cus_y] + bcost[next_x][next_y] - '0') {

				continue;

			}

			s.erase( entry(next_x, next_y) );

			dd[next_x][next_y] = dd[cus_x][cus_y] + bcost[next_x][next_y] - '0';

			s.insert( entry(next_x, next_y) );

		}

	}

	maxC = 0;

	for (i = 0; i < rows; i++) {

		for (j = 0; j < cols; j++) {

			maxC = max(maxC, dd[i][j]);

		}

	}

	return maxC;

}