ZOJ 1649 Rescue (BFS)

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http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1649

做得蛋疼！！！

要注意以下几点：

cost[i,j]表示friend到[i,j]的代价
每次队列最前元素的周围四个元素，都要计算其代价，即使已经被访问过
如果出队列的元素是angle，不需要立即break，因为队列中后续元素可能有更小代价

#include <iostream>#include <vector>#include <algorithm>#include <queue>#include <string.h>#include <stdio.h>using namespace std;#define MAXINT 0x7FFFFFFFint dx[] = {0, 1, 0, -1};int dy[] = {-1, 0, 1, 0};int main(){int N, M;char matrix[200][200];int cost[200][200];while (cin >> N >> M){for (int i = 0; i < N; i++){for (int j = 0; j < M; j++)cin >> matrix[i][j];getchar();}int min = MAXINT;for (int i = 0; i < N; i++)for (int j = 0; j < M; j++)if (matrix[i][j] == 'r') // a friend{// initializefor (int k = 0; k < N; k++)for (int h = 0; h < M; h++)cost[k][h] = MAXINT;// add first onecost[i][j] = 0;queue< pair<int, int> > q;q.push(make_pair<int, int>(i, j));while (!q.empty()){pair<int, int> c = q.front();q.pop();if (matrix[c.first][c.second] == 'a'){if (cost[c.first][c.second] < min)min = cost[c.first][c.second];}else{for (int k = 0; k < 4; k++){int x = c.first + dx[k];int y = c.second + dy[k];if (0 <= x && x < N && 0 <= y && y < M  && matrix[x][y] != '#'){if (cost[x][y] == MAXINT ||(matrix[c.first][c.second] == 'x' && cost[x][y] > cost[c.first][c.second] + 2) ||(matrix[c.first][c.second] != 'x' && cost[x][y] > cost[c.first][c.second] + 1)){q.push(make_pair<int, int>(x, y));}if (matrix[c.first][c.second] == 'x' && cost[x][y] > cost[c.first][c.second] + 2){cost[x][y] = cost[c.first][c.second] + 2;}else if (matrix[c.first][c.second] != 'x' && cost[x][y] > cost[c.first][c.second] + 1){cost[x][y] = cost[c.first][c.second] + 1;}}}}}}if (min == MAXINT)cout << "Poor ANGEL has to stay in the prison all his life." << endl;elsecout << min << endl;}}