The Shortest Path in Nya Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4872 Accepted Submission(s): 1122
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 <= 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 li (1 <= li <= N), which is the layer of ith 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.
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 li (1 <= li <= N), which is the layer of ith 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.
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
题意: 有n个地点，m条边，每条边耗费w，每个点又属于一些集合，相邻的集合之间的点可以直接通行，耗费c。现在求1到n的最少费用。
思路:可以把集合当做一个点，从集合到集合内的点费用为0，点到相邻的集合费用为c，这样就既保证了不用暴力枚举相邻集合里面的每个点，有保证相邻集合里的点的最短路（表达能力不好 -。-#！）。
就是这张图的意思：
然后一遍dij+heap就好。
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<time.h>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = ;
struct node
{
int to;
int val;
int next;
}edge[MAXN * ];
int pre[MAXN*],vis[MAXN*],dis[MAXN*],n,m,k,c,ind;
int num[MAXN],ok[MAXN];
void add(int x,int y,int z)
{
edge[ind].to = y;
edge[ind].val = z;
edge[ind].next = pre[x];
pre[x] = ind ++;
}
struct Node
{
int p;
ll val;
Node(){}
Node(int fp,int fval):p(fp),val(fval){}
friend bool operator < (Node fa,Node fb){
return fa.val > fb.val;
}
};
void dij(int s)
{
priority_queue<Node>fq;
for(int i = ; i <= k; i++){
dis[i] = INF;
vis[i] = ;
}
dis[s] = ;
fq.push(Node(s,));
while(!fq.empty()){
Node tp = fq.top();
fq.pop();
vis[tp.p] = ;
for(int i = pre[tp.p]; i != -; i = edge[i].next){
int t = edge[i].to;
if(!vis[t] && dis[t] > dis[tp.p] + edge[i].val){
dis[t] = dis[tp.p] + edge[i].val;
fq.push(Node(t,dis[t]));
}
}
}
}
int main()
{
int t,ff = ;
scanf("%d",&t);
while(t--){
memset(num,,sizeof(num));
memset(ok,,sizeof(ok));
scanf("%d%d%d",&n,&m,&c);
int x;
ind = ;
memset(pre,-,sizeof(pre));
for(int i = ; i <= n; i++){
scanf("%d",&x);
ok[x] = ;
num[i] = x;
}
for(int i = ; i <= n; i++){//点与集合之间连边
add(n+num[i],i,);
if(num[i] > )add(i,n+num[i]-,c);
if(num[i] < n)add(i,n+num[i]+,c);
}
int y,z;
for(int i = ; i <= m; i++){
scanf("%d%d%d",&x,&y,&z);
add(x,y,z);
add(y,x,z);
}
//cout<<k<<endl;
k = * n;
dij();
printf("Case #%d: ",++ff);
if(dis[n] >= INF){
printf("-1\n");
}
else {
printf("%d\n",dis[n]);
}
}
return ;
}