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题意:给一张图,有火源,有障碍物,剩下的是道路,火源在下一分钟能够让上下左右四个方向的道路也着火,告诉人的位置,问最短时间能逃出去的时间是多少;
思路:一个bfs用来求出所有的火源能蔓延到的地方,另一个bfs求最短路,本来我以为bfs只能求最短路;
超级源:有多个源头的时候,就把所有的都压进去,每个源头能蔓延到的地方都看作是新的源头,继续压进去,每个只能蔓延4个方向,直到空;
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cmath> 5 #include <queue> 6 #include <algorithm> 7 #include <vector> 8 #include <cstring> 9 #include <stack> 10 #include <cctype> 11 #include <utility> 12 #include <map> 13 #include <string> 14 #include <climits> 15 #include <set> 16 #include <string> 17 #include <sstream> 18 #include <utility> 19 #include <ctime> 20 using namespace std; 21 const int MAXN(1010); 22 const int MAXM(1000010); 23 int fire[MAXN][MAXN]; 24 char g[MAXN][MAXN]; 25 bool vis[MAXN][MAXN]; 26 int fx[MAXM], fy[MAXM]; 27 int R, C, SX, SY; 28 struct NODE 29 { 30 int x, y, val; 31 NODE() 32 {} 33 NODE(int tx, int ty, int td): x(tx), y(ty), val(td) 34 {} 35 }; 36 37 int ok(int nx,int ny) 38 { 39 if(nx >= 1 && nx <= R && ny >= 1 && ny <= C ) 40 return 1; 41 return 0; 42 } 43 int move_x[4] = {0, -1, 0, 1}; 44 int move_y[4] = {-1, 0, 1, 0}; 45 46 void bfs1(int n)///多个起点,超级源 47 { 48 queue<NODE> q; 49 memset(fire+1, -1, sizeof(fire)); 50 for(int i = 0; i < n; ++i)///起点都压进去 51 { 52 q.push(NODE(fx[i], fy[i], 0)); 53 fire[fx[i]][fy[i]] = 0; 54 } 55 NODE cur; 56 while(!q.empty()) 57 { 58 cur = q.front();///从一个起点开始,然后四个方向扩散 59 q.pop(); 60 for(int i = 0; i < 4; ++i) 61 { 62 int nx = cur.x+move_x[i], ny = cur.y+move_y[i]; 63 if(ok(nx,ny)&& g[nx][ny] != ‘#‘ && fire[nx][ny] == -1) 64 { 65 fire[nx][ny] = cur.val+1; 66 q.push(NODE(nx, ny, cur.val+1)); 67 ///重新做一个新的起点 68 } 69 } 70 } 71 } 72 73 int bfs2() 74 { 75 queue<NODE> q; 76 memset(vis+1, false, sizeof(vis[0])*R); 77 q.push(NODE(SX, SY, 0)); 78 vis[SX][SY] = true; 79 NODE cur; 80 while(!q.empty()) 81 { 82 cur = q.front(); 83 q.pop(); 84 if(cur.x == 1 || cur.y == 1 || cur.x == R || cur.y == C) 85 return cur.val+1; 86 for(int i = 0; i < 4; ++i) 87 { 88 int nx = cur.x+move_x[i], ny = cur.y+move_y[i]; 89 if(ok(nx,ny)&& g[nx][ny] != ‘#‘ && !vis[nx][ny] && (fire[nx][ny] == -1 || cur.val+1 < fire[nx][ny])) 90 { 91 vis[nx][ny] = true; 92 q.push(NODE(nx, ny, cur.val+1)); 93 } 94 } 95 } 96 return -1; 97 } 98 99 int main() 100 { 101 int T; 102 scanf("%d", &T); 103 while(T--) 104 { 105 scanf("%d%d", &R, &C); 106 int count = 0; 107 for(int i = 1; i <= R; ++i) 108 { 109 scanf("%s", g[i]+1); 110 for(int j = 1; j <= C; ++j) 111 { 112 if(g[i][j] == ‘J‘) 113 { 114 SX = i; 115 SY = j; 116 } 117 if(g[i][j] == ‘F‘) 118 { 119 fx[count] = i; 120 fy[count++] = j; 121 } 122 } 123 } 124 bfs1(count); 125 int ans = bfs2(); 126 if(ans == -1) 127 printf("IMPOSSIBLE\n"); 128 else 129 printf("%d\n", ans); 130 } 131 return 0; 132 }
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原文地址:http://www.cnblogs.com/ACMERY/p/4495970.html