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G++ 用pair过不了为啥

Posted by shadofight1v3 at 2015-09-08 21:16:12 on Problem 2243
以下代码是用C++过的,用G++就是过不去
#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
using namespace std;
int vis[10][10];
int dir[8][2] = {{2,1},{2,-1},{-2,-1},{-2,1},{1,2},{1,-2},{-1,-2},{-1,2}};
int sx, sy, ex, ey;
int BFS() {
	pair<int, int> st = make_pair(sx, sy);
	queue< pair<int, int> > Q;
	Q.push(st);
	vis[sx][sy] = 1;
	if(sx == ex && sy == ey) return 0;
	int x, y, t;
	int step = 0;
	while(!Q.empty()) {
		t = Q.size(); step++;
		while(t--) {
			st = Q.front(); Q.pop();
			for(int i = 0; i < 8; i++) {
				x = st.first + dir[i][0];
				y = st.second + dir[i][1];
				if(x>=0&&x<8&&y>=0&&y<8&&!vis[x][y]) {
					vis[x][y] = 1;
					if(x == ex && y == ey) {
						return step;
					}
					Q.push(make_pair(x, y));
				}
			}
		}
	}
	return 0;
}
int main()
{
	char str1[5], str2[5];
	while(scanf("%s%s", str1, str2) != EOF) {
		memset(vis, 0, sizeof(vis));
		sx = str1[0] - 'a'; sy = str1[1] - '0' - 1;
		ex = str2[0] - 'a'; ey = str2[1] - '0' - 1;
		//cout << sx << sy << " " << ex << ey << endl;
		int step = BFS();
		printf("To get from %s to %s takes %d knight moves.\n", str1, str2, step);
	}
	return 0;
}

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