标签：时间 first algorithm pair field art mat 题目 cond
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1180
TLE n次。。
注意两点：1，S.T这三种位置是可以停留一秒的。即在没有路可走的时候可以停留一秒。
2， bfs宽搜应该使用优先队列， 并且vis标记加在将next节点push到队列中的时候。
然后就是奇偶判断什么的就可以了。
代码：
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 #include <cstring> 6 #include <cassert> 7 #include <cstdio> 8 #include <cctype> 9 #include <vector> 10 #include <string> 11 #include <queue> 12 #include <stack> 13 #include <cmath> 14 #include <map> 15 #include <set> 16 using namespace std; 17 #define rep(i,a,n) for (int i=a;i<n;i++) 18 #define per(i,a,n) for (int i=n-1;i>=a;i--) 19 #define pb push_back 20 #define mp make_pair 21 #define all(x) (x).begin(),(x).end() 22 #define fi first 23 #define se second 24 #define SZ(x) ((int)(x).size()) 25 typedef vector<int> VI; 26 typedef long long ll; 27 typedef pair<int, int> PII; 28 const ll mod = 1000000007; 29 ll powmod(ll a, ll b) { ll res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res*a%mod; a = a*a%mod; }return res; } 30 // head 31 const int inf = 0x3f3f3f3f; 32 #define maxn 22 33 int iadd[] = {0, 0, 1, -1}, jadd[] = {1, -1, 0, 0}; 34 struct node{ 35 int i, j, stp; 36 node(int ii, int jj, int sstp): i(ii), j(jj), stp(sstp) {} 37 bool operator < (const node &t) const { 38 return stp > t.stp; 39 } 40 }; 41 char maze[maxn][maxn]; 42 int n, m; 43 int vis[maxn][maxn]; 44 bool flag = false; 45 priority_queue<node> q; 46 47 int bfs(int stai, int staj, int dsti, int dstj){ 48 if(n == 0 || m == 0) return 0; 49 if(n == 1 && m == 1) return 0; 50 flag = false; 51 memset(vis, 0, sizeof(vis)); 52 while(!q.empty()) q.pop(); 53 vis[stai][staj] = 1; 54 q.push(node(stai, staj, 0)); 55 56 int maxqsize = 0; 57 while(!q.empty()){ 58 // maxqsize = max(maxqsize, (int)q.size()); 59 node tmn = q.top(); 60 q.pop(); 61 // q.push(node(tmn.i, tmn.j, tmn.stp + 1)); 62 if(tmn.i == dsti && tmn.j == dstj){ 63 // printf("-->%d\n", maxqsize); 64 return tmn.stp; 65 } 66 //printf("##\n%d\n##\n", tmn.stp); 67 bool upd = false; 68 for(int i = 0; i < 4; i++){ 69 int nxti = tmn.i + iadd[i]; 70 int nxtj = tmn.j + jadd[i]; 71 if(nxti < 0 || nxti >= n || nxtj < 0 || nxtj >= m) 72 continue; 73 if(vis[nxti][nxtj] || maze[nxti][nxtj] == ‘*‘) 74 continue; 75 if(maze[nxti][nxtj] == ‘.‘ || maze[nxti][nxtj] == ‘T‘){ 76 upd = true; 77 vis[nxti][nxtj] = 1; 78 q.push(node(nxti, nxtj, tmn.stp + 1)); 79 } 80 else{ 81 //horizontal 82 if(i < 2){ 83 //next 84 if(nxtj < m - 1 && i == 0){ 85 if(maze[nxti][nxtj + 1] != ‘*‘){ 86 if(tmn.stp & 1 && maze[nxti][nxtj] == ‘|‘){ 87 upd = true; 88 vis[nxti][nxtj + 1] = 1; 89 q.push(node(nxti, nxtj + 1, tmn.stp + 1)); 90 } 91 //even 92 else if(!(tmn.stp & 1) && maze[nxti][nxtj] == ‘-‘){ 93 upd = true; 94 vis[nxti][nxtj + 1] = 1; 95 q.push(node(nxti, nxtj + 1, tmn.stp + 1)); 96 } 97 98 } 99 } 100 //previous 101 else if(nxtj > 0 && i == 1){ 102 if(maze[nxti][nxtj - 1] != ‘*‘){ 103 if(tmn.stp & 1 && maze[nxti][nxtj] == ‘|‘){ 104 upd = true; 105 vis[nxti][nxtj - 1] = 1; 106 q.push(node(nxti, nxtj - 1, tmn.stp + 1)); 107 } 108 //even 109 else if(!(tmn.stp & 1) && maze[nxti][nxtj] == ‘-‘){ 110 upd = true; 111 vis[nxti][nxtj - 1] = 1; 112 q.push(node(nxti, nxtj - 1, tmn.stp + 1)); 113 } 114 } 115 } 116 } 117 //vertical 118 else{ 119 if(maze[nxti + 1][nxtj] != ‘*‘){ 120 if(nxti < n - 1 && i == 2){ 121 if(tmn.stp & 1 && maze[nxti][nxtj] == ‘-‘){ 122 upd = true; 123 vis[nxti + 1][nxtj] = 1; 124 q.push(node(nxti + 1, nxtj, tmn.stp + 1)); 125 } 126 //even 127 else if(!(tmn.stp & 1) && maze[nxti][nxtj] == ‘|‘){ 128 upd = true; 129 vis[nxti + 1][nxtj] = 1; 130 q.push(node(nxti + 1, nxtj, tmn.stp + 1)); 131 } 132 } 133 } 134 //previous 135 else if(nxti > 0 && i == 3){ 136 if(maze[nxti - 1][nxtj] != ‘*‘){ 137 if(tmn.stp & 1 && maze[nxti][nxtj] == ‘-‘){ 138 upd = true; 139 vis[nxti - 1][nxtj] = 1; 140 q.push(node(nxti - 1, nxtj, tmn.stp + 1)); 141 } 142 //even 143 if(!(tmn.stp & 1) && maze[nxti][nxtj] == ‘|‘){ 144 upd = true; 145 vis[nxti - 1][nxtj] = 1; 146 q.push(node(nxti - 1, nxtj, tmn.stp + 1)); 147 } 148 } 149 } 150 } 151 } 152 } 153 if(!upd){ 154 q.push(node(tmn.i, tmn.j, tmn.stp + 1)); 155 } 156 } 157 158 return 0; 159 } 160 161 162 int main(){ 163 while(scanf("%d %d", &n, &m) != EOF){ 164 memset(maze, 0, sizeof(maze)); 165 int si, sj, ti, tj; 166 for(int i = 0; i < n; i++){ 167 for(int j = 0; j < m; j++){ 168 scanf(" %c", &maze[i][j]); 169 if(maze[i][j] == ‘S‘) 170 si = i, sj = j; 171 if(maze[i][j] == ‘T‘) 172 ti = i, tj = j; 173 } 174 } 175 printf("%d\n", bfs(si, sj, ti, tj)); 176 } 177 } 178 179 180 181 182 183 /* 184 5 5 185 **..T 186 **.*. 187 ..|.. 188 .*.*. 189 S.... 190 5 5 191 S|.|. 192 |.|*| 193 .|... 194 |***| 195 **.|T 196 3 4 197 S|.| 198 -T-. 199 .|.. 200 20 20 201 S.|.|.|.|.|.|.|.|.|. 202 .|.|.|.|.|.|.|.|.|.| 203 |.|.|.|.|.|.|.|.|.|. 204 .|.|.|.|.|.|.|.|.|.| 205 |.|.|.|.|.|.|.|.|.|. 206 .|.|.|.|.|.|.|.|.|.| 207 |.|.|.|.|.|.|.|.|.|. 208 .|.|.|.|.|.|.|.|.|.| 209 |.|.|.|.|.|.|.|.|.|. 210 .|.|.|.|.|.|.|.|.|.| 211 |.|.|.|.|.|.|.|.|.|. 212 .|.|.|.|.|.|.|.|.|.| 213 |.|.|.|.|.|.|.|.|.|. 214 .|.|.|.|.|.|.|.|.|.| 215 |.|.|.|.|.|.|.|.|.|. 216 .|.|.|.|.|.|.|.|.|.| 217 |.|.|.|.|.|.|.|.|.|. 218 .|.|.|.|.|.|.|.|.|.| 219 |.|.|.|.|.|.|.|.|.|. 220 .|.|.|.|.|.|.|.|.|.T 221 5 6 222 **..|T 223 **.*.. 224 ..|.-. 225 .*.*.. 226 S..... 227 20 20 228 **..**.*|.**.*.|T... 229 **.*.|....|.....*.*. 230 ..|....|...*.*.**.** 231 .*.*.**.*.**.*.|.... 232 |.....*.*...|....|.. 233 **..**.*|..*.*.**.*. 234 **.*.|......|....|.. 235 ..|....|..**.*.|.... 236 .*.*.**.*.|.....*.*. 237 |.....*.*..*|..*.*.. 238 **..**.*|.**.*.|.... 239 **.*.|**.*|..*.*.**. 240 **.*.*......|....|.. 241 ..|....|...*.*.**.** 242 .*.*.**.*.**.*.|.... 243 |.....*.*...|....|.. 244 **......|...*.*.*... 245 ..|....|..**.*.|.... 246 .*.*.**.*.|.....*.*. 247 S.....*.*..*|..*.*.. 248 */
题目：
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 14593 Accepted Submission(s): 3736
标签：时间 first algorithm pair field art mat 题目 cond
原文地址：http://www.cnblogs.com/bolderic/p/6869593.html
