HDU4725(KB4-P SPFA+LLL+SLF优化)

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7981    Accepted Submission(s): 1794

Problem Description

This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just solo hay que cambiar un poco el algoritmo. If you do not understand a word of this paragraph, just move on.
The Nya graph is an undirected graph with "layers". Each node in the graph belongs to a layer, there are N nodes in total.
You can move from any node in layer x to any node in layer x + 1, with cost C, since the roads are bi-directional, moving from layer x + 1 to layer x is also allowed with the same cost.
Besides, there are M extra edges, each connecting a pair of node u and v, with cost w.
Help us calculate the shortest path from node 1 to node N.

 

Input

The first line has a number T (T <= 20) , indicating the number of test cases.
For each test case, first line has three numbers N, M (0 <= N, M <= 10⁵) and C(1 <= C <= 10³), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers l_i (1 <= l_i <= N), which is the layer of i^th node belong to.
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 10⁴), which means there is an extra edge, connecting a pair of node u and v, with cost w.

 

Output

For test case X, output "Case #X: " first, then output the minimum cost moving from node 1 to node N.
If there are no solutions, output -1.

 

Sample Input

2
3 3 3
1 3 2
1 2 1
2 3 1
1 3 3

3 3 3
1 3 2
1 2 2
2 3 2
1 3 4

 

Sample Output

Case #1: 2
Case #2: 3

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

SLF：Small Label First 策略，设要加入的节点是j，队首元素为i，若dist(j)<dist(i)，则将j插入队首，否则插入队尾。

LLL：Large Label Last 策略，设队首元素为i，队列中所有dist值的平均值为x，若dist(i)>x则将i插入到队尾，查找下一元素，直到找到某一i使得dist(i)<=x，则将i出队进行松弛操作。

 

建图：

相邻层之间建无向边，权值为c

第i层向该层上的节点建有向边，权值为0

节点i向其所在层的相邻层建有向边，权值为c。

然后额外m条无向边。

 //2017-08-29
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <queue>
 #include <stack>

 using namespace std;

 const int N = ;
 const int M = ;
 const int INF = 0x3f3f3f3f;

 int head[N], tot;
 struct Edge{
     int to, next, w;
 }edge[M];

 void init(){
     tot = ;
     memset(head, -, sizeof(head));
 }

 void add_edge(int u, int v, int w){
     edge[tot].w = w;
     edge[tot].to = v;
     edge[tot].next = head[u];
     head[u] = tot++;
 }

 int n, m, c;
 bool vis[N];
 int dis[N], cnt[N];

 bool spfa(int s, int n){
     memset(vis, , sizeof(vis));
     memset(dis, INF, sizeof(dis));
     memset(cnt, , sizeof(cnt));
     vis[s] = ;
     dis[s] = ;
     cnt[s] = ;
     deque<int> dq;
     dq.push_back(s);
     int sum = , len = ;
     while(!dq.empty()){
         // LLL 优化
         while(dis[dq.front()]*len > sum){
             dq.push_back(dq.front());
             dq.pop_front();
         }
         int u = dq.front();
         sum -= dis[u];
         len--;
         dq.pop_front();
         vis[u] = ;
         for(int i = head[u]; i != -; i = edge[i].next){
             int v = edge[i].to;
             if(dis[v] > dis[u] + edge[i].w){
                 dis[v] = dis[u] + edge[i].w;
                 if(!vis[v]){
                     vis[v] = ;
                     // SLF 优化
                     if(!dq.empty() && dis[v] < dis[dq.front()])
                       dq.push_front(v);
                     else dq.push_back(v);
                     sum += dis[v];
                     len++;
                     if(++cnt[v] > n)return false;
                 }
             }
         }
     }
     return true;
 }

 bool book[N];
 int layer[N];

 int main()
 {
     //freopen("inputP.txt", "r", stdin);
     int T, kase = ;
     scanf("%d", &T);
     while(T--){
         scanf("%d%d%d", &n, &m, &c);
         init();
         memset(book, , sizeof(book));
         for(int i = ; i <= n; i++){
             scanf("%d", &layer[i]);
             book[layer[i]] = ;
         }
         for(int i = ; i <= n; i++){
             add_edge(n+layer[i], i, );
             if(layer[i] >  && book[layer[i]-])
                   add_edge(i, n+layer[i]-, c);
             if(layer[i] < n && book[layer[i]+])
                   add_edge(i, n+layer[i]+, c);
         }
         for(int i = ; i < N; i++)
           if(book[i] && book[i+]){
               add_edge(i+n, i++n, c);
               add_edge(i++n, i+n, c);
           }
         int u, v, w;
         while(m--){
             scanf("%d%d%d", &u, &v, &w);
             add_edge(u, v, w);
             add_edge(v, u, w);
         }
         spfa(, n<<);
         printf("Case #%d: %d\n", ++kase, dis[n]==INF?-:dis[n]);
     }

     return ;
 }