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 = 300100;
const int M = 2500100;
const int INF = 0x3f3f3f3f;

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

void init(){
    tot = 0;
    memset(head, -1, 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, 0, sizeof(vis));
    memset(dis, INF, sizeof(dis));
    memset(cnt, 0, sizeof(cnt));
    vis[s] = 1;
    dis[s] = 0;
    cnt[s] = 1;
    deque<int> dq;
    dq.push_back(s);
    int sum = 0, len = 1;
    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] = 0;
        for(int i = head[u]; i != -1; 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] = 1;
                    // 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 = 0;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d%d", &n, &m, &c);
        init();
        memset(book, 0, sizeof(book));
        for(int i = 1; i <= n; i++){
            scanf("%d", &layer[i]);
            book[layer[i]] = 1;
        }
        for(int i = 1; i <= n; i++){
            add_edge(n+layer[i], i, 0);
            if(layer[i] > 1 && book[layer[i]-1])
                  add_edge(i, n+layer[i]-1, c);
            if(layer[i] < n && book[layer[i]+1])
                  add_edge(i, n+layer[i]+1, c);
        }
        for(int i = 1; i < N; i++)
          if(book[i] && book[i+1]){
              add_edge(i+n, i+1+n, c);
              add_edge(i+1+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(1, n<<1);
        printf("Case #%d: %d
", ++kase, dis[n]==INF?-1:dis[n]);
    }

    return 0;
}