HDU - 4725 The Shortest Path in Nya Graph （最短路）
The Shortest Path in Nya Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10675 Accepted Submission(s): 2330
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.
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.
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.
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
Source
2013 ACM/ICPC Asia Regional Online —— Warmup2
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题意：n个点分布在n层，每一层的任意一个点都可以到达相邻层的任意一点，花费为0，还有m条边，连接这n个点，边的权值为花费，求出从1到n的最小距离
思路：这道题一开始我想简单了，我直接在松弛的时候做了些处理，结果T了，仔细一想我那种做法其实就是把相邻两层所有的点都互相加了边，不T才怪，正解是把第i层也看做是一个点，为第i+n个点，每一层到属于该层的点加一条单向边，权值为0，如果相邻两层都存在点的话，点到相邻层添加一条单向边，权值为c，然后跑最短路
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int MAXN = 200005;
const int INF = 0x3f3f3f3f;
struct edge
{
int to,next;
int w;
}e[MAXN * 10];
struct node
{
int id;
int dis;
node(){}
node(int _id,int _dis){id = _id,dis = _dis;}
bool operator <(const struct node & a) const{
return dis > a.dis;
}
};
int head[MAXN],cnt;
int cen[MAXN];
void addEdge(int u,int v,ll w)
{
e[cnt].to = v;
e[cnt].w = w;
e[cnt].next = head[u];
head[u] = cnt++;
}
int belong[MAXN];
int dis[MAXN];
int vis[MAXN];
int n,m,c;
void Dijkstra()
{
struct node now;
for(int i = 1; i <= 2 * n; i++) {
dis[i] = INF;
vis[i] = 0;
}
priority_queue q;
dis[1] = 0;
q.push(node(1,0));
while(!q.empty()) {
now = q.top();
q.pop();
int u = now.id,v;
if(vis[u]) continue;
vis[u] = 1;
for(int i = head[u]; i != -1; i = e[i].next) {
v = e[i].to;
if(dis[v] > dis[u] + e[i].w) {
dis[v] = dis[u] + e[i].w;
q.push(node(v,dis[v]));
}
}
}
}
int main(void)
{
int Case = 0;
int T;
int u,v,w;
scanf("%d",&T);
while(T--) {
Case++;
cnt = 0;
memset(head,-1,sizeof(head));
memset(cen,0,sizeof(cen));
scanf("%d%d%d",&n,&m,&c);
for(int i = 1; i <= n; i++) {
scanf("%d",&belong[i]);
//每一层到属于该层的点加一条单向边
addEdge(belong[i] + n,i,0);
cen[belong[i]]++;
}
//如果相邻两层都存在点的话，点到相邻层添加一条单向边
for(int i = 1; i <= n; i++) {
if(belong[i] > 1 && cen[belong[i] - 1] > 0) addEdge(i,belong[i] - 1 + n,c);
if(belong[i] < n && cen[belong[i] + 1] > 0) addEdge(i,belong[i] + 1 + n,c);
}
for(int i = 1; i <= m; i++) {
scanf("%d%d%d",&u,&v,&w);
addEdge(u,v,w);
addEdge(v,u,w);
}
Dijkstra();
if(dis[n] == INF) printf("Case #%d: -1\n",Case);
else printf("Case #%d: %d\n",Case,dis[n]);
}
return 0;
}