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HDU 1072 Nightmare BFS
其实就是多加了一个引爆时间的限制条件,反正n,m给的很小,直接记录3维状态,之后就很随意了。
#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;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 15;const int WALL = 0;const int EMPTY = 1;const int STR = 2;const int TAR = 3;const int RESET = 4;const int dx[] = {0,0,1,-1}, dy[] = {1,-1,0,0};int mp[maxn][maxn], st[maxn][maxn][7];int n,m,sx,sy,ex,ey;void bfs() { queue<int> qx,qy,qt; qx.push(sx); qy.push(sy); qt.push(6); st[sx][sy][6] = 0; while(!qx.empty()) { int x = qx.front(), y = qy.front(), t = qt.front(); int nowt = st[x][y][t]; qx.pop(); qy.pop(); qt.pop(); for(int i = 0;i < 4;i++) { int nx = x + dx[i], ny = y + dy[i]; int nt = mp[nx][ny] == RESET ? 6 : t - 1; if(t - 1 > 0 && mp[nx][ny] != WALL) { if(nowt + 1 < st[nx][ny][nt]) { st[nx][ny][nt] = nowt + 1; qx.push(nx); qy.push(ny); qt.push(nt); } } } }}int main() { int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); memset(mp,0,sizeof(mp)); memset(st,0x3f,sizeof(st)); int inf = st[0][0][0],ans = inf; for(int i = 1;i <= n;i++) { for(int j = 1;j <= m;j++) { scanf("%d",&mp[i][j]); if(mp[i][j] == STR) { sx = i; sy = j; } if(mp[i][j] == TAR) { ex = i; ey = j; } } } bfs(); for(int i = 1;i < 6;i++) ans = min(ans,st[ex][ey][i]); if(ans >= inf) puts("-1"); else printf("%d\n",ans); } return 0;}
HDU 1072 Nightmare BFS
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