题目链接
http://poj.org/problem?id=3984
思路 因为要找最短路 用BFS
而且 每一次 往下一层搜 要记录当前状态 之前走的步的坐标
最后 找到最短路后 输出坐标就可以了
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a) memset(a, 0, sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; typedef pair<string, int> psi; typedef pair<string, string> pss; const double PI = acos(-1); const double E = exp(1); const double eps = 1e-30; const int INF = 0x3f3f3f3f; const int maxn = 5e4 + 5; const int MOD = 1e9 + 7; int G[5][5]; int v[5][5]; int Move[4][2] { -1, 0, 1, 0, 0,-1, 0, 1, }; struct Node { int x, y; vector <pii> ans; }tmp; vector <pii> ans; queue <Node> q; bool ok(int x, int y) { if (x < 0 || x >= 5 || y < 0 || y >= 5 || v[x][y] || G[x][y]) return false; return true; } void bfs() { tmp.x = 0; tmp.y = 0; tmp.ans.pb(pii(0, 0)); v[tmp.x][tmp.y] = 1; q.push(tmp); while (!q.empty()) { int x = q.front().x; int y = q.front().y; ans = q.front().ans; q.pop(); if (x == 4 && y == 4) return; for (int i = 0; i < 4; i++) { tmp.x = x + Move[i][0]; tmp.y = y + Move[i][1]; if (ok(tmp.x, tmp.y)) { tmp.ans = ans; tmp.ans.pb(pii(tmp.x, tmp.y)); q.push(tmp); tmp.ans.pop_back(); v[tmp.x][tmp.y] = 1; } } } } int main() { CLR(v); for (int i = 0; i < 5; i++) { for (int j = 0; j < 5; j++) scanf("%d", &G[i][j]); } bfs(); vector <pii>::iterator it; for (it = ans.begin(); it != ans.end(); it++) { printf("(%d, %d)\n", (*it).first, (*it).second); } }转载于:https://www.cnblogs.com/Dup4/p/9433136.html
