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
#include <cstdio>
#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,c; void init()
{
memset(head, , sizeof(head));
ecnt = ;
} 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));
}
}
}
return ;
}
/*有n个点m条无向边，每个点有一个层次，相邻层次的点可以移动，花费为C，问1~n的最小花费。
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层的点连一条费用为0的边，
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。*/
int main()
{
int T;
scanf("%d",&T);
for(int t = ; t <= T; ++t)
{
scanf("%d %d %d",&n,&m,&c);
init();
for(int i = ; i <= n; ++i)
{
scanf("%d",&lay[i]);
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);
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]);
}
return ;
}
最短路（HDU4725）