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HDU4725TheShortestPathinNyaGraph(最短路径)(2013ACM/ICPCAsiaRegionalOnline――Warmup2)

DescriptionThisisaveryeasyproblem,yourtaskisjustcalculateelcaminomascortoe

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):

 1 #include 
2 #include
3 #include
4 #include
5 #include
6 using namespace std;
7 typedef pair<int, int> PII;
8
9 const int MAXN = 300010;
10 const int MAXE = MAXN * 2;
11
12 int head[MAXN];
13 int to[MAXE], next[MAXE], cost[MAXE];
14 int n, m, ecnt, lcnt, c;
15
16 void init() {
17 memset(head, 0, sizeof(head));
18 ecnt = lcnt = 1;
19 }
20
21 inline void add_edge(int u, int v, int c) {
22 to[ecnt] = v; cost[ecnt] = c; next[ecnt] = head[u]; head[u] = ecnt++;
23 }
24
25 int dis[MAXN];
26 int lay[MAXN];
27 bool vis[MAXN];
28
29 void Dijkstra(int st, int ed) {
30 memset(dis, 0x7f, sizeof(dis));
31 memset(vis, 0, sizeof(vis));
32 priority_queue que; que.push(make_pair(0, st));
33 dis[st] = 0;
34 while(!que.empty()) {
35 int u = que.top().second; que.pop();
36 if(vis[u]) continue;
37 if(u == ed) return ;
38 vis[u] = true;
39 for(int p = head[u]; p; p = next[p]) {
40 int &v = to[p];
41 if(dis[v] > dis[u] + cost[p]) {
42 dis[v] = dis[u] + cost[p];
43 que.push(make_pair(-dis[v], v));
44 }
45 }
46 }
47 }
48
49 int T;
50
51 inline int readint() {
52 char c = getchar();
53 while(!isdigit(c)) c = getchar();
54 int ret = 0;
55 while(isdigit(c)) ret = ret * 10 + c - '0', c = getchar();
56 return ret;
57 }
58
59 int main() {
60 T = readint();
61 for(int t = 1; t <= T; ++t) {
62 n = readint(), m = readint(), c = readint();
63 init();
64 for(int i = 1; i <= n; ++i) {
65 lay[i] = readint();
66 add_edge(i, n + 2 * lay[i] - 1, 0);
67 add_edge(n + 2 * lay[i], i, 0);
68 }
69 for(int i = 1; i i) {
70 add_edge(n + 2 * i - 1, n + 2 * (i + 1), c);
71 add_edge(n + 2 * (i + 1) - 1, n + 2 * i, c);
72 }
73 int u, v, w;
74 while(m--) {
75 //scanf("%d%d%d", &u, &v, &w);
76 u = readint(), v = readint(), w = readint();
77 add_edge(u, v, w);
78 add_edge(v, u, w);
79 }
80 Dijkstra(1, n);
81 if(dis[n] == 0x7f7f7f7f) dis[n] = -1;
82 printf("Case #%d: %d\n", t, dis[n]);
83 }
84 }
View Code

 


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