HDU 1885 Key Task

发布时间:2016-12-6 8:56:25 编辑:www.fx114.net 分享查询网我要评论
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bfs。100*100的地图上,有4种钥匙,可能有多个或没有出口,可能有钥匙没门,也可能有门没钥匙。压缩一下,三维数组判重。 #pragma comment(linker, "/STACK:1024000000,1024000000") #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<vector> #include<queue> #include<cmath> ///LOOP #define REP(i, n) for(int i = 0; i < n; i++) #define FF(i, a, b) for(int i = a; i < b; i++) #define FFF(i, a, b) for(int i = a; i <= b; i++) #define FD(i, a, b) for(int i = a - 1; i >= b; i--) #define FDD(i, a, b) for(int i = a; i >= b; i--) ///INPUT #define RI(n) scanf("%d", &n) #define RII(n, m) scanf("%d%d", &n, &m) #define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k) #define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p) #define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q) #define RFI(n) scanf("%lf", &n) #define RFII(n, m) scanf("%lf%lf", &n, &m) #define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k) #define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p) #define RS(s) scanf("%s", s) ///OUTPUT #define PN printf("\n") #define PI(n) printf("%d\n", n) #define PIS(n) printf("%d ", n) #define PS(s) printf("%s\n", s) #define PSS(s) printf("%s ", n) #define PC(n) printf("Case %d: ", n) ///OTHER #define PB(x) push_back(x) #define CLR(a, b) memset(a, b, sizeof(a)) #define CPY(a, b) memcpy(a, b, sizeof(b)) #define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}} #define lson l, m, rt << 1 #define rson m + 1, r, rt << 1 | 1 using namespace std; typedef long long LL; typedef pair<int, int> P; const int MOD = 9901; const int INFI = 1e9 * 2; const LL LINFI = 1e17; const double eps = 1e-6; const double pi = acos(-1.0); const int N = 111; const int M = 1111; const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1}; struct point { int x, y, k, t; point(){}; point(int a, int b, int c, int d){x = a, y = b, k = c, t = d;}; }p; int n, m, mm[255] = {0}; char s[N][N]; bool vis[N][N][16]; queue<point> q; bool check(int x, int y) { return x < n && y < m && x >= 0 && y >= 0; } void bfs() { int tx, ty, key; while(!q.empty()) { p = q.front(); q.pop(); REP(i, 4) { tx = p.x + move[i][0]; ty = p.y + move[i][1]; if(check(tx, ty) && !vis[tx][ty][p.k] && s[tx][ty] != '#') { if(s[tx][ty] == 'X') { printf("Escape possible in %d steps.\n", p.t + 1); return ; } else if(s[tx][ty] == '.') { vis[tx][ty][p.k] = 1; q.push(point(tx, ty, p.k, p.t + 1)); } else if(s[tx][ty] >= 'A' && s[tx][ty] <= 'Z') { if(p.k & mm[s[tx][ty]]) { vis[tx][ty][p.k] = 1; q.push(point(tx, ty, p.k, p.t + 1)); } } else { key = (p.k | mm[s[tx][ty]]); vis[tx][ty][p.k] = 1; q.push(point(tx, ty, key, p.t + 1)); } } } } puts("The poor student is trapped!"); } int main() { //freopen("input.txt", "r", stdin); mm['B'] = mm['b'] = 1; mm['Y'] = mm['y'] = 2; mm['R'] = mm['r'] = 4; mm['G'] = mm['g'] = 8; while(RII(n, m), n + m) { CLR(vis, 0); while(!q.empty())q.pop(); REP(i, n) { RS(s[i]); REP(j, m) { if(s[i][j] == '*') { s[i][j] = '.'; vis[i][j][0] = 1; q.push(point(i, j, 0, 0)); } } } bfs(); } return 0; }

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