problem
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 M 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.
Examples
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
题目大意：
给定n个点，在一般的基础上给每个点一个维度，也就是多了一个参数表示所在的层次，现在已知：
1、X层的点可以通过增加权值C到达X-1层或X+1层中的任何一个点
2、给定m个已经形成的路径
将层转换理解为虚点，N开双倍，2e5。
这里要注意虚点操作时，过渡的边应该这么建立：
1、因为X层的点 i 通过增加权值C才能到达X-1层或X+1层
所以sz[i].push_back(edge(t+n-1,c));和sz[i].push_back(edge(t+n+1,c));
2、假设点 j 已经通过权值C到达X层，所以他可以到达X层所有点，即X层表示的虚点到X层的所有点的距离为0。所以sz[t+n].push_back(edge(i,0));
不要搞反了，这点很重要，因为涉及到你是否理解这题的虚点操作。
下面这个代码我为了省事，用cin，实际题面会T，用scanf或输入优化就可以过了。
#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn=2e5+7;
struct edge{
int to,w;
edge(int a,int b){
to=a; w=b; }
};
vector<edge>sz[maxn];
int d[maxn],T,n,m,c;
bool vis[maxn];
void spfa(){
memset(vis,0,sizeof(vis)); memset(d,inf,sizeof(d));
d[1]=0; vis[1]=1;
queue<int>q; q.push(1);
while(!q.empty()){
int t=q.front(); q.pop(); vis[t]=0;
for(int i=0;i<sz[t].size();i++){
int y=sz[t][i].to;
int x=sz[t][i].w;
if(d[y]>d[t]+x){
d[y]=d[t]+x;
if(!vis[y]){
vis[y]=1;
q.push(y);
}
}
}
}
}
int main(){
cin>>T;
for(int tt=1;tt<=T;tt++){
cin>>n>>m>>c;
for(int i=1;i<=2*n;i++) sz[i].clear();
for(int i=1;i<=n;i++){
int t; cin>>t;
sz[t+n].push_back(edge(i,0));
if(t>1) sz[i].push_back(edge(t+n-1,c));
if(t<n) sz[i].push_back(edge(t+n+1,c));
}
for(int i=1;i<=m;i++){
int a,b,c;
cin>>a>>b>>c;
sz[a].push_back(edge(b,c));
sz[b].push_back(edge(a,c));
}
spfa();
// for(int i=1;i<=n;i++) cout<<d[i]<<' ';
// putchar('\n');
if(d[n]==inf) d[n]=-1;
printf("Case #%d: %d\n",tt,d[n]);
}
}
/*
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
Case #1: 2
Case #2: 3
*/
可以看其他人的题解：链接，还有图解