题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=103921#problem/J
题意和思路见代码注释吧~~ 粗暴的两个bfs。。还真是可以~~
只用一个bfs 应该就是可以的。但是我没有想开~~也没研究~~
1 /* 2 大意是一个地图,开始的时候有一个人,然后还有几个位置着火了,想问人能否逃出,如果能,最短时间是多少。 3 开始想先以人为起点,bfs,计算出人到达每个位置的时间。然后以火苗为起点 计算出火最早蔓延的时间 比较一下。 4 有些麻烦,好像都放在一个队列里就可以吗?好像是不行的, 5 */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <queue> 11 #define maxn 1000000 12 using namespace std; 13 14 int n, m; 15 char mp[2100][2100]; 16 bool visPer[2100][2100], visFire[2100][2100]; 17 int stepPer[2100][2100]; 18 int stepFire[2100][2100]; 19 20 21 struct Node{ 22 int x; 23 int y; 24 }; 25 26 queue<Node>que; 27 queue<Node>queper; 28 29 int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1}; 30 31 bool checkPer(Node a) { 32 if (a.x >=0 && a.x < n && a.y >= 0 && a.y < m && !visPer[a.x][a.y] && mp[a.x][a.y] != '#') { 33 return true; 34 } 35 return false; 36 } 37 38 bool checkFire(Node a) { 39 if (a.x >=0 && a.x < n && a.y >= 0 && a.y < m && !visFire[a.x][a.y] && mp[a.x][a.y] != '#') { 40 return true; 41 } 42 return false; 43 } 44 45 void dfsPer() { 46 while(!queper.empty()) { 47 Node now = queper.front(); 48 queper.pop(); 49 50 for (int i=0; i<4; ++i) { 51 Node temp; 52 temp.x = now.x + dir[i][0]; 53 temp.y = now.y + dir[i][1]; 54 // cout << checkPer(temp) << "===" << stepPer[now.x][now.y] + 1 << endl; 55 if (checkPer(temp) && stepFire[temp.x][temp.y] > stepPer[now.x][now.y] + 1) { 56 queper.push(temp); 57 visPer[temp.x][temp.y] = 1; 58 stepPer[temp.x][temp.y] = stepPer[now.x][now.y] + 1; 59 } 60 } 61 } 62 } 63 64 65 66 void dfsFire() { 67 while(!que.empty()) { 68 Node now = que.front(); 69 que.pop(); 70 71 for (int i=0; i<4; ++i) { 72 Node temp; 73 temp.x = now.x + dir[i][0]; 74 temp.y = now.y + dir[i][1]; 75 if (checkFire(temp)) { 76 que.push(temp); 77 visFire[temp.x][temp.y] = 1; 78 stepFire[temp.x][temp.y] = stepFire[now.x][now.y] + 1; 79 } 80 } 81 } 82 } 83 84 int main() { 85 int t; 86 cin >> t; 87 while(t--) { 88 cin >> n >> m; 89 for (int i=0; i<n; ++i) { 90 for (int j=0; j<m; ++j) { 91 stepFire[i][j] = maxn; 92 stepPer[i][j] = maxn; 93 } 94 } 95 while(!que.empty()) { 96 que.pop(); 97 } 98 while(!queper.empty()) { 99 queper.pop(); 100 } 101 memset(visPer, 0, sizeof(visPer)); 102 memset(visFire, 0, sizeof(visFire)); 103 104 Node per; 105 for (int i=0; i<n; ++i) { 106 for (int j=0; j<m; ++j) { 107 cin >> mp[i][j]; 108 if (mp[i][j] == 'F') { 109 Node now; 110 now.x = i, now.y = j; 111 que.push(now); 112 stepFire[i][j] = 1; 113 visFire[i][j] = 1; 114 } 115 else if (mp[i][j] == 'J') { 116 per.x = i, per.y = j; 117 visPer[i][j] = 1; 118 stepPer[i][j] = 1; 119 queper.push(per); 120 } 121 } 122 } 123 dfsFire(); 124 dfsPer(); 125 126 int stepmin = maxn; 127 for (int i=0; i<n; ++i) { 128 stepmin = min(stepmin, stepPer[i][0]); 129 stepmin = min(stepmin, stepPer[i][m-1]); 130 } 131 132 for (int i=0; i<m; ++i) { 133 stepmin = min(stepmin, stepPer[0][i]); 134 stepmin = min(stepmin, stepPer[n-1][i]); 135 } 136 137 if (stepmin == maxn) { 138 cout << "IMPOSSIBLE "; 139 } 140 else cout << stepmin << endl; 141 } 142 return 0; 143 }