poj1122 FDNY to the Rescue!(dij+反向建图+输出路径)

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define CLR(a,b) memset((a),(b),sizeof((a)))
 5 using namespace std;
 6 
 7 const int inf = 0x3f3f3f3f;
 8 const int N = 21;
 9 int n, fire;
10 int g[N][N];//邻接矩阵存图
11 int s[N], d[N];
12 int path[N];//表示v0到vi最短路径顶点vi的前一个顶点序号
13 int shortest[N];//存储最短路径上的各个顶点序号
14 struct node{
15     int u, v, t;
16 }a[N];
17 int cmp(node a, node b){
18     return a.t < b.t;
19 }
20 void dij(){
21     int i, j, k;
22     CLR(s, 0);
23     for(i = 1; i <= n; ++i){
24         d[i] = g[fire][i];
25         if(i != fire)
26             path[i] = fire;
27         else
28             path[i] = -1;
29     }
30     s[fire] = 1;
31     d[fire] = 0;
32     for(i = 0; i < n-1; ++i){
33         int mi = inf, u = 0;
34         for(j = 1; j <= n ; ++j){
35             if(!s[j] && d[j] < mi){
36                 u = j; mi = d[j];
37             }
38         }
39         s[u] = 1;
40         for(k = 1; k <= n; ++k){
41             if(!s[k] && d[u] + g[u][k] < d[k]){
42                 d[k] = d[u] + g[u][k];
43                 path[k] = u;
44             }
45         }
46     }
47 }
48 int main(){
49     int i, j, x, cnt;
50     scanf("%d", &n);
51     for(i = 1; i <= n; ++i){//边反向存储
52         for(j = 1; j <= n; ++j){
53             scanf("%d", &x);
54             if(x == -1)
55                 g[j][i] = inf;
56             else
57                 g[j][i] = x;
58         }
59     }
60     scanf("%d", &fire);
61     dij();
62     cnt = 0;
63     while(scanf("%d", &x) == 1){
64         a[cnt].u = x;
65         a[cnt].v = fire;
66         a[cnt++].t = d[x];
67     }
68     sort(a, a+cnt, cmp);
69     printf("Org\tDest\tTime\tPath\n");
70     for(i = 0; i < cnt; ++i){
71         printf("%d\t%d\t%d", a[i].u, a[i].v, a[i].t);
72         CLR(shortest, 0);
73         int k = 0;//表示shortest数组中最后一个元素的下标
74         shortest[k] = a[i].u;
75         while(path[shortest[k]] != -1){
76             k++;
77             shortest[k] = path[shortest[k-1]];
78         }
79         for(j = 0; j <= k; ++j) //倒向追踪，但是反向建图，因此正向输出
80             printf("\t%d", shortest[j]);
81         puts("");
82     }
83     return 0;
84 }