HDU 2612 Find a way(bfs)
2015-05-18 21:02
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解析:
求2个点到KFC的距离之和,使其最小,可用2次BFS,分别求出2个点到各个KFC的最短距离,然后找出和最小的即可。AC代码
[code]#include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <cstdlib> using namespace std; typedef long long ll; const int INF = 0x3f3f3f3f; const int N = 205; const int dx[] = {-1, 0, 1, 0}; const int dy[] = { 0, 1, 0,-1}; char grid ; int dp [2]; bool vis [2]; struct Node { int x, y, step; Node() {} Node(int _x, int _y, int _step) { x = _x, y = _y, step = _step; } }poi[2]; int n, m; bool judge(int x, int y) { if(x < 0 || x >= n || y < 0 || y >= m || grid[x][y] == '#') return true; return false; } void bfs(int sx, int sy, int ch) { queue<Node> que; dp[sx][sy][ch] = 0; vis[sx][sy][ch] = true; que.push(Node(sx, sy, 0)); int x, y, step; while(!que.empty()) { Node front = que.front(); que.pop(); for(int i = 0; i < 4; i++) { x = front.x + dx[i]; y = front.y + dy[i]; step = front.step + 1; if(judge(x, y) || vis[x][y][ch]) continue; vis[x][y][ch] = true; que.push(Node(x, y, step)); if(grid[x][y] == '@') { dp[x][y][ch] = step; } } } } int main() { while(scanf("%d%d", &n, &m) != EOF) { memset(dp, 0, sizeof(dp)); memset(vis, false, sizeof(vis)); for(int i = 0; i < n; i++) { scanf("%s", grid[i]); for(int j = 0; j < m; j++) { if(grid[i][j] == 'Y') poi[0].x = i, poi[0].y = j; else if(grid[i][j] == 'M') poi[1].x = i, poi[1].y = j; } } for(int i = 0; i <= 1; i++) { bfs(poi[i].x, poi[i].y, i); } int ans = INF; for(int i = 0; i < n; i++) { for(int j = 0; j < m; j++) { if(grid[i][j] == '@' && vis[i][j][0] && vis[i][j][1]) ans = min(ans, dp[i][j][0] + dp[i][j][1]); } } printf("%d\n", ans*11); } return 0; }
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