The Shortest Path in Nya Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9444 Accepted Submission(s): 2098
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:给你点和每个点所在的层,层和层之间移动有花费
方法1:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=1e5+5;
int n,m,c,l;
int head[maxn],point[maxn],vis[maxn],d[maxn];
vector<int> dc[maxn];
struct ac{
int to,next,value;
}eg[2*maxn];
void add(int u,int v,int w){
eg[l].to=v;
eg[l].value=w;
eg[l].next=head[u];
head[u]=l++;
}
struct ak{
bool operator()(const int &a,const int &b){ //优先队列排序规则
return d[a]>d[b];
}
};
void dijkstra(int star){
memset(vis,0,sizeof(vis));
memset(d,INF,sizeof(d));
priority_queue<int, vector<int>, ak> q;
d[star]=0;
q.push(star);
while(!q.empty()){
int x=q.top();q.pop();
for(int k=head[x];k!=-1;k=eg[k].next){
int y=eg[k].to;
if(d[y]>d[x]+eg[k].value){
d[y]=d[x]+eg[k].value;
q.push(y);
}
}
if(point[x]<n&&!vis[point[x]+1]){ //判断要不往上走一层
vis[point[x]+1]=1;
for(int i=0;i<dc[point[x]+1].size();i++){
int y=dc[point[x]+1][i];
if(d[y]>d[x]+c){
d[y]=d[x]+c;
q.push(y);
}
}
}
if(point[x]>1 &&!vis[point[x]-1]){ //判断要不要往下走一层
vis[point[x]-1]=1;
for(int i=0;i<dc[point[x]-1].size();i++){
int y=dc[point[x]-1][i];
if(d[y]>d[x]+c){
d[y]=d[x]+c;
q.push(y);
}
}
}
}
}
int main(){
int t,f=0;
scanf("%d",&t);
while(t--){
l=0;
scanf("%d %d %d",&n,&m,&c);
memset(head,-1,sizeof(head));
memset(point,0,sizeof(point));
int u,v,w;
for(int i=1;i<=n;i++){
scanf("%d",&u);
point[i]=u; //第i个点在第u层
dc[u].push_back(i); //第u层有i点
}
for(int i=0;i<m;i++){
scanf("%d %d %d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
dijkstra(1);
f++;
printf("Case #%d: ", f);
if(d[n]==INF||n==0) printf("-1\n");
else printf("%d\n",d[n]);
for(int i = 1; i <= n; ++i) dc[i].clear();
}
return 0;
}
方法2:
把点和层相连,点和层的距离为0,把层当成一个新的点
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 5 * 1e5;
int n, m, c;
int pos[maxn];
vector<int> laye[maxn];
int head[maxn];
int d[maxn];
bool vis[maxn];
struct edge{
int to, nex, cost;
}eg[maxn];
int cnt = 0;
void add(int u,int v,int cost) {
eg[cnt].to = v;
eg[cnt].nex = head[u];
eg[cnt].cost = cost;
head[u] = cnt++;
}
struct Rule {
bool operator () (int a,int b) const {
return d[a] > d[b];
}
};
void dijkstra(int s) {
memset(d, inf, sizeof(d));
priority_queue<int, vector<int>, Rule> q;
d[s] = 0;
q.push(s);
while(!q.empty()) {
int u = q.top(); q.pop();
for(int k = head[u]; k != -1; k = eg[k].nex) {
int v = eg[k].to;
if(d[v] > d[u] + eg[k].cost) {
// printf("%d %d %d\n", u, v,eg[k].cost);
d[v] = d[u] + eg[k].cost;
q.push(v);
}
}
}
}
int main()
{
int t;
scanf("%d", &t);
for(int cas = 1; cas <= t; ++cas) {
cnt = 0;
memset(d, inf, sizeof(d));
memset(head, -1, sizeof(head));
memset(pos, 0, sizeof(pos));
scanf("%d %d %d", &n, &m, &c);
int u, v, cost;
for(int i = 1; i <= n; ++i) {
scanf("%d", &u);
pos[i] = u;
add(i, n + 2 * u - 1, 0);//层内的点到该层距离为0
add(n + 2 * u, i, 0);
vis[u] = true;
}
for(int i = 1; i < n; ++i) {
if(vis[i] && vis[i + 1]) {
add(n + 2 * i - 1, n + 2 * (i + 1), c);//相邻层之间的距离为c
add(n + 2 * (i + 1) - 1, n + 2 * i, c);
}
}
for(int i =0; i < m; ++i) {
scanf("%d %d %d", &u, &v, &cost);
add(u, v, cost);
add(v, u, cost);
}
dijkstra(1);
if(d[n] == inf || n == 0)
printf("Case #%d: -1\n", cas);
else
printf("Case #%d: %d\n", cas, d[n]);
}
return 0;
}