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题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2671
题意:一个人在树林里要到出口,同时有几个火点在扩展。问这个人能不能跑出树林,最少多少步能跑出去。
先扩展人,再扩展火。在同一个队列里扩展,所以要记下接下来要扩展几个人或者火点。
1 #include <algorithm> 2 #include <iostream> 3 #include <iomanip> 4 #include <cstring> 5 #include <climits> 6 #include <complex> 7 #include <fstream> 8 #include <cassert> 9 #include <cstdio> 10 #include <bitset> 11 #include <vector> 12 #include <deque> 13 #include <queue> 14 #include <stack> 15 #include <ctime> 16 #include <set> 17 #include <map> 18 #include <cmath> 19 20 using namespace std; 21 22 typedef pair<int, int> PII; 23 typedef struct P { 24 int x; 25 int y; 26 int s; 27 P() {} 28 P(int xx, int yy, int ss) : x(xx), y(yy), s(ss) {} 29 }P; 30 31 const int inf = 0x7f7f7f; 32 const int maxn = 1010; 33 const int dx[4] = {0, 0, 1, -1}; 34 const int dy[4] = {1, -1, 0, 0}; 35 36 int r, c, jx, jy; 37 vector<PII> fir; 38 int ans; 39 int vis[maxn][maxn]; 40 char G[maxn][maxn]; 41 42 inline int min(int x, int y) { 43 return x < y ? x : y; 44 } 45 46 bool judge(int x, int y) { 47 if(!vis[x][y] && G[x][y] == ‘.‘ && x >= 0 && x < r && y >= 0 && y < c) { 48 return 1; 49 } 50 return 0; 51 } 52 53 bool success(int x, int y) { 54 if(G[x][y] == ‘J‘ && 55 (x == 0 || x == r - 1 || y == 0 || y == c - 1)) { 56 return 1; 57 } 58 return 0; 59 } 60 61 void init() { 62 memset(vis, 0, sizeof(vis)); 63 memset(G, 0, sizeof(G)); 64 ans = inf; 65 fir.clear(); 66 } 67 68 void bfs() { 69 P q[maxn*maxn]; 70 int front = 0, tail = 0; 71 int fcnt = 0, jcnt = 1, tmp = 0; 72 for(int i = 0; i < fir.size(); i++) { 73 q[tail++] = P(fir[i].first, fir[i].second, 0); 74 vis[fir[i].first][fir[i].second] = 1; 75 fcnt++; 76 } 77 q[tail++] = P(jx, jy, 0); vis[jx][jy] = 1; 78 while(front < tail) { 79 for(int k = 0; k < fcnt; k++) { 80 P f = q[front++]; 81 for(int i = 0; i < 4; i++) { 82 int xx = f.x + dx[i]; 83 int yy = f.y + dy[i]; 84 if(judge(xx, yy) && G[xx][yy] != ‘F‘) { 85 vis[xx][yy] = 1; 86 if(G[xx][yy] != ‘J‘) { 87 G[xx][yy] = ‘F‘; 88 } 89 q[tail++] = P(xx, yy, f.s+1); tmp++; 90 } 91 } 92 } 93 fcnt = tmp; tmp = 0; 94 for(int k = 0; k < jcnt; k++) { 95 P j = q[front++]; 96 if(success(j.x, j.y)) { 97 ans = min(ans, j.s); 98 } 99 for(int i = 0; i < 4 ; i++) { 100 int xx = j.x + dx[i]; 101 int yy = j.y + dy[i]; 102 if(judge(xx, yy) && G[xx][yy] != ‘F‘) { 103 vis[xx][yy] = 1; 104 if(G[xx][yy] != ‘F‘) { 105 G[xx][yy] = ‘J‘; 106 } 107 q[tail++] = P(xx, yy, j.s+1); tmp++; 108 } 109 } 110 } 111 jcnt = tmp; tmp = 0; 112 } 113 } 114 115 int main() { 116 int T; 117 scanf("%d", &T); 118 while(T--) { 119 init(); 120 scanf("%d %d", &r, &c); 121 for(int i = 0; i < r; i++) { 122 scanf("%s", G[i]); 123 } 124 for(int i = 0; i < r; i++) { 125 for(int j = 0; j < c; j++) { 126 if(G[i][j] == ‘J‘) { 127 jx = i; jy = j; 128 } 129 if(G[i][j] == ‘F‘) { 130 fir.push_back(PII(i, j)); 131 } 132 } 133 } 134 bfs(); 135 if(ans == inf) { 136 printf("IMPOSSIBLE\n"); 137 } 138 else { 139 printf("%d\n", ans+1); 140 } 141 } 142 return 0; 143 }
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原文地址:http://www.cnblogs.com/vincentX/p/5874560.html