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HDU 5040 Instrusive BFS
如果题意明确了的话就是一个简单bfs。。。。。。
用优先队列搞一下还是很快的。
#include <cstdio>#include <cstring>#include <iostream>#include <map>#include <set>#include <vector>#include <string>#include <queue>#include <deque>#include <bitset>#include <list>#include <cstdlib>#include <climits>#include <cmath>#include <ctime>#include <algorithm>#include <stack>#include <sstream>#include <numeric>#include <fstream>#include <functional> using namespace std; #define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int, int> PII;typedef pair<double, double> PDD;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 505;const int dx[] = {-1, 0, 1, 0, 0};const int dy[] = {0, 1, 0, -1, 0};map<char, int> dmp;char mp[maxn][maxn];int n, sx, sy, ex, ey;void init() { dmp[‘S‘] = 0; dmp[‘W‘] = 1; dmp[‘N‘] = 2; dmp[‘E‘] = 3;}struct Node { int dist, x, y; Node(int x, int y, int dist): dist(dist), x(x), y(y) {} bool operator < (const Node &node) const { if(dist == node.dist) { return abs(x - ex) + abs(y + ey) > abs(node.x - ex) + abs(node.y - ey); } return dist > node.dist; }};bool cvis[maxn][maxn];bool can() { memset(cvis, 0, sizeof(cvis)); queue<PII> q; q.push(MP(sx, sy)); while(!q.empty()) { PII now = q.front(); q.pop(); int x = now.first, y = now.second; if(x == ex && y == ey) break; for(int i = 0; i < 4; i++) { int nx = x + dx[i], ny = y + dy[i]; if(mp[nx][ny] != ‘#‘ && !cvis[nx][ny]) { q.push(MP(nx, ny)); cvis[nx][ny] = true; } } } return cvis[ex][ey];}bool vis[maxn][maxn][4];int getstate(int x, int y, int nowtime) { if(mp[x][y] == ‘#‘) return -1; if(mp[x][y] != ‘.‘) return 1; for(int i = 0; i < 4; i++) { int nx = x + dx[i], ny = y + dy[i]; if(mp[nx][ny] == ‘.‘ || mp[nx][ny] == ‘#‘) continue; int nd = (dmp[mp[nx][ny]] + nowtime) % 4; if(nd == i) return 1; } return 0;}void solve() { if(!can()) { puts("-1"); return; } memset(vis, 0, sizeof(vis)); priority_queue<Node> q; q.push(Node(sx, sy, 0)); vis[sx][sy][0] = true; while(!q.empty()) { Node now = q.top(); q.pop(); int x = now.x, y = now.y, dist = now.dist; if(x == ex && y == ey) { printf("%d\n", dist); return; } for(int i = 0; i < 5; i++) { if(i == 4) { if(!vis[x][y][(dist + 1) % 4]) { vis[x][y][(dist + 1) % 4] = true; q.push(Node(x, y, dist + 1)); } continue; } int nx = x + dx[i], ny = y + dy[i]; int nowstate = getstate(x, y, dist), nxtstate = getstate(nx, ny, dist); if(nxtstate == -1) continue; if(nowstate == 0 && nxtstate == 0) { if(vis[nx][ny][(dist + 1) % 4]) continue; vis[nx][ny][(dist + 1) % 4] = true; q.push(Node(nx, ny, dist + 1)); } else if(!vis[nx][ny][(dist + 3) % 4]) { vis[nx][ny][(dist + 3) % 4] = true; q.push(Node(nx, ny, dist + 3)); } } } puts("-1");}int main() { init(); int T; scanf("%d", &T); for(int kase = 1; kase <= T; kase++) { scanf("%d", &n); memset(mp, ‘#‘, sizeof(mp)); for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { scanf(" %c", &mp[i][j]); if(mp[i][j] == ‘M‘) { sx = i; sy = j; mp[i][j] = ‘.‘; } if(mp[i][j] == ‘T‘) { ex = i; ey = j; mp[i][j] = ‘.‘; } } } printf("Case #%d: ", kase); solve(); } return 0;}
HDU 5040 Instrusive BFS
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