Easy Graph Problem?(一道简单图论题?) bfs + 优先队列

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <queue>
  4 using namespace std;
  5 typedef long long LL;
  6 const int INF = 0X5FFFFF;
  7 const int MS = 505;
  8 
  9 struct node {
 10     int x, y, dir, cost;
 11     bool operator < (const node & a) const {
 12         return cost > a.cost;
 13     }
 14 };
 15 int pic[MS][MS];
 16 int dir[4][2] = {0,1,1,0,0,-1,-1,0};
 17 int flag[MS][MS][4];
 18 int minv[MS][MS];
 19 
 20 int n, m, r1, c1, r2, c2;
 21 int kase = 1;
 22 node s,t;
 23 
 24 void bfs1() {
 25     minv[r1][c1] = pic[r1][c1];
 26     s.x = r1;
 27     s.y = c1;
 28     s.cost = pic[r1][c1];
 29     priority_queue<node> que;
 30     que.push(s);
 31     while (!que.empty()) {
 32         s = que.top();
 33         que.pop();
 34         for (int i = 0; i < 4; i++) {
 35             int x = s.x + dir[i][0];
 36             int y = s.y + dir[i][1];
 37             if (x > 0 && x <= n && y > 0 && y <= m && pic[x][y]) {
 38                 t.x = x;
 39                 t.y = y;
 40                 t.cost = s.cost + pic[x][y];
 41                 if (x == r2 && y == c2) {
 42                     printf(" %d", t.cost);
 43                     return ;
 44                 }
 45                 if (t.cost < minv[x][y]) {
 46                     minv[x][y] = t.cost;
 47                     que.push(t);
 48                 }
 49             }
 50         }
 51     }
 52     printf(" -1");
 53 }
 54 
 55 void bfs2() {
 56     s.x = r1;
 57     s.y = c1;
 58     s.cost = pic[r1][c1];
 59     priority_queue<node> que;
 60     for (int i = 0; i < 4; i++) {
 61         flag[r1][c1][i] = 1;
 62         int x = s.x + dir[i][0];
 63         int y = s.y + dir[i][1];
 64         if (x > 0 && x <= n && y > 0 && y <= m && pic[x][y]) {
 65             flag[x][y][i] = 1;
 66             t.x = x;
 67             t.y = y;
 68             t.cost = s.cost + pic[x][y];
 69             t.dir = i;
 70             que.push(t);
 71         }
 72     }
 73     while (!que.empty()) {
 74         s = que.top();
 75         que.pop();
 76         for (int i = 0; i < 4; i++) {
 77             if (i == s.dir)
 78                 continue;
 79             t.x = s.x + dir[i][0];
 80             t.y = s.y + dir[i][1];
 81             if (t.x > 0 && t.x <= n && t.y > 0 && t.y <= m && pic[t.x][t.y]) {
 82                 t.cost = s.cost + pic[t.x][t.y];
 83                 t.dir = i;
 84                 if (t.x == r2 && t.y == c2) {
 85                     printf(" %d",t.cost);
 86                     return ;
 87                 }
 88                 if (flag[t.x][t.y][t.dir] == 0) {
 89                     flag[t.x][t.y][t.dir] = 1;
 90                     que.push(t);
 91                 }
 92             }
 93 
 94         }
 95     }
 96     printf(" -1");
 97 }
 98 
 99 int main() {
100     char str[MS];
101     while (scanf("%d%d%d%d%d%d", &n, &m, &r1, &c1, &r2, &c2) != EOF) {
102         for (int i  = 1; i <= n; i++)
103             for (int j = 1; j <= m; j++) {
104                 scanf("%s", str);
105                 if (str[0] == ‘*‘) {
106                     pic[i][j] = 0;
107                 }
108                 else {
109                     sscanf(str, "%d", &pic[i][j]);  // 初始化图
110                 }
111                 minv[i][j] = INF;   //  (i,j) 最小花费初始化
112                 for (int k = 0; k < 4; k++)
113                     flag[i][j][k] = 0;  // (i,j) 四个方向初始化
114             }
115         printf("Case %d:",kase++);
116         bfs1();
117         bfs2();
118         printf("\n");
119     }
120     return 0;
121 }