The Shortest Path in Nya Graph
The Shortest Path in Nya Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9374 Accepted Submission(s): 2073
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.
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.
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.
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
唉~~~~~~~ 难受
就是一道点与点之间的最短路
这道题主要分为两个部分，存点和dijkstra
因为如果按常规方法存为超时，所以要把点拆分来存（看网上百度到的）
因为是点与点之间的最短路所以可以用 ：同一层的点之间的距离为零 ，相邻层的点距离为c
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先把每一层的点指向本层 ， 距离为零 点->层
在把每两个相邻层 用层指向点 层 -> 点
================================
这样就可以通过层为媒介把点指向点
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include 迪杰斯特拉原代码
void dijkstra()
{
memset(dis,INF,sizeof(dis));
memset(flag,0,sizeof(flag));
dis[1]=0;
for(int i=1;i<=n1;i++){
int u=-1,min=INF;
for(int j=1;j<=n1;j++){
if(!flag[j]&&min>dis[j]){
u=j;
min=dis[j];
}
}
if(u==-1)
return ;
flag[u]=1;
for(int j=0;j优化后的迪杰斯特拉+bfs
void dijkstra()
{
memset(dis,INF,sizeof(dis));
memset(flag,0,sizeof(flag));
dis[1]=0;
priority_queuequ;
ac temp_1;
temp_1.v=1;
temp_1.c=0;
qu.push(temp_1);
while(!qu.empty()){
ac temp_2=qu.top();
qu.pop();
int v=temp_2.v;
if(dis[v]dis[v]+temp_3.c){
temp_3.c=dis[temp_3.v]=dis[v]+temp_3.c;//因为que是优先队列，所以如果要把temp_3入队，要把temp_3.c的值更新（挠头） 一开始没想到，在大佬的帮助下，才改正。。
qu.push(temp_3);
}
}
}
} 唉，就这样吧，如果各位大佬发现有错的地方可以在评论里指出来，以便小弟及时改正·······