HDU 4725 The Shortest Path in Nya Graph（最短路径）（2013 ACM/ICPC Asia Regional Online ―― Warmup2）

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 <= 10 ⁵) and C(1 <= C <= 10 ³), 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 <= 10 ⁴), 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.

题目大意：有n个点m条无向边，然后每个点有一个层次，相邻层次的点可以移动（同层次不可以），花费为C，问1~n的最小花费。

思路：我试过类似于每层之间的点都连一条边（不是$O(n^2)$那种很挫的方法，不断TLE……好吧换思路……每层新建两个点a、b，i层的点到ai连一条费用为0的边，bi到i层的点连一条边，然后相邻的层分别从ai到bj连一条边，费用为C。跑Dijkstra+heap可AC。

PS：至于最初的思路为啥会TLE我就不想说了……有兴趣自己动脑吧……

代码（343MS）：

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 using namespace std;
 typedef pair<int, int> PII;

 const int MAXN = ;
 const int MAXE = MAXN * ;

 int head[MAXN];
 int to[MAXE], next[MAXE], cost[MAXE];
 int n, m, ecnt, lcnt, c;

 void init() {
     memset(head, , sizeof(head));
     ecnt = lcnt = ;
 }

 inline void add_edge(int u, int v, int c) {
     to[ecnt] = v; cost[ecnt] = c; next[ecnt] = head[u]; head[u] = ecnt++;
 }

 int dis[MAXN];
 int lay[MAXN];
 bool vis[MAXN];

 void Dijkstra(int st, int ed) {
     memset(dis, 0x7f, sizeof(dis));
     memset(vis, , sizeof(vis));
     priority_queue<PII> que; que.push(make_pair(, st));
     dis[st] = ;
     while(!que.empty()) {
         int u = que.top().second; que.pop();
         if(vis[u]) continue;
         if(u == ed) return ;
         vis[u] = true;
         for(int p = head[u]; p; p = next[p]) {
             int &v = to[p];
             if(dis[v] > dis[u] + cost[p]) {
                 dis[v] = dis[u] + cost[p];
                 que.push(make_pair(-dis[v], v));
             }
         }
     }
 }

 int T;

 inline int readint() {
     char c = getchar();
     while(!isdigit(c)) c = getchar();
     int ret = ;
     while(isdigit(c)) ret = ret *  + c - '', c = getchar();
     return ret;
 }

 int main() {
     T = readint();
     for(int t = ; t <= T; ++t) {
         n = readint(), m = readint(), c = readint();
         init();
         for(int i = ; i <= n; ++i) {
             lay[i] = readint();
             add_edge(i, n +  * lay[i] - , );
             add_edge(n +  * lay[i], i, );
         }
         for(int i = ; i < n; ++i) {
             add_edge(n +  * i - , n +  * (i + ), c);
             add_edge(n +  * (i + ) - , n +  * i, c);
         }
         int u, v, w;
         while(m--) {
             //scanf("%d%d%d", &u, &v, &w);
             u = readint(), v = readint(), w = readint();
             add_edge(u, v, w);
             add_edge(v, u, w);
         }
         Dijkstra(, n);
         if(dis[n] == 0x7f7f7f7f) dis[n] = -;
         printf("Case #%d: %d\n", t, dis[n]);
     }
 }