题意：给一个方阵表示一个湖每个格子代表那个位置的风向，顺风走不要花费否则花费为1。给你起点重点问你最小花费是多少？
思路：这道题写的四不像，先spfa写了一下T，然后bfs又T。接着加了个优先队列就过了。。。主要就是类似于spfa dis数组记录到那点的最小花费。然后只要能更新就入队。
代码如下：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cmath> 5 #include <cstring> 6 #include <algorithm> 7 #include <queue> 8 #include <stack> 9 #include <vector> 10 #include <set> 11 #include <map> 12 #define PB(a) push_back(a) 13 14 using namespace std; 15 16 typedef long long ll; 17 typedef pair<int ,int> pii; 18 typedef pair<unsigned int, unsigned int> puu; 19 typedef pair<int ,double> pid; 20 typedef pair<ll, int> pli; 21 typedef pair<int, ll> pil; 22 23 const int INF = 0x3f3f3f3f; 24 const double eps = 1e-6; 25 const int LEN = 1002; 26 int Map[LEN][LEN], n, m, dis[LEN][LEN]; 27 int xx[] = {-1,-1, 0, 1, 1, 1, 0,-1}; 28 int yy[] = { 0, 1, 1, 1, 0,-1,-1,-1}; 29 struct P {int x, y, st;}; 30 struct cmp 31 { 32 bool operator() (P a, P b){return a.st>b.st;} 33 }; 34 35 P temp; 36 37 int bfs(int x, int y, int ex, int ey) 38 { 39 temp.x = x;temp.y = y;temp.st = 0; 40 priority_queue<P, vector<P>, cmp> q; 41 int ret = INF; 42 q.push(temp); 43 register int tx, ty, i; 44 memset(dis, 0x3f, sizeof dis); 45 dis[x][y] = 0; 46 while(!q.empty()){ 47 P vex = q.top();q.pop(); 48 if(vex.x==ex && vex.y==ey) {return vex.st;} 49 for(i=0; i<8; i++){ 50 tx = vex.x+xx[i]; 51 ty = vex.y+yy[i]; 52 temp.st = vex.st+(i==Map[vex.x][vex.y]?0:1); 53 if(tx>=0 && tx<n && ty>=0 && ty<m && dis[tx][ty] > temp.st){ 54 temp.x = tx;temp.y = ty; 55 dis[tx][ty] = temp.st; 56 q.push(temp); 57 } 58 } 59 } 60 } 61 62 void read(int &ret) 63 { 64 char c; 65 while((c = getchar())<'0' || c>'9'); 66 ret = 0; 67 while(c>='0' && c<='9'){ 68 ret = ret*10+(c-'0'); 69 c = getchar(); 70 } 71 } 72 73 int main() 74 { 75 // freopen("in.txt", "r", stdin); 76 77 int q, sx, sy, ex, ey, ans; 78 while(scanf("%d%d", &n, &m)!=EOF){ 79 getchar(); 80 for(int i=0; i<n; i++){ 81 for(int j=0; j<m; j++){ 82 Map[i][j] = getchar()-'0'; 83 } 84 getchar(); 85 } 86 read(q); 87 for(int i=0; i<q; i++){ 88 read(sx);read(sy);read(ex);read(ey); 89 ans = bfs(sx-1, sy-1, ex-1, ey-1); 90 printf("%d ", ans); 91 } 92 } 93 return 0; 94 }