HDU - 4725
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
题目描述
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
题意
给定一幅有层次的线路图，第一行输入
，
，
表示
个几点和
条边，每两层之间通过的费用是
。
意思就是，除了走给定的路之外，我们还可以选择穿透层次，花费
的费用走到上一层或者下一层的任意一个节点，而不去走题目给定的边。
然后第二行
个数就是表示第
个节点的层号，然后
行是描述边的。
我们可以考虑，怎么样去维护层与层之间的连通，是他们相互通过花费为
呢？我们考虑每层新添加两个虚拟节点，一个只出不进，一个只进不出。
只出不进的结点可以连着上下两层只进不出的结点。如图所示
由于 spfa 超时了，所以后来改成了迪杰斯特拉。
代码
#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int INF = 0x3fffffff;
const int maxn = 5e5 + 10;
int path[maxn], head[maxn], lay[maxn], T, n, m, c, u, v, x, l, cnt, ca;
bool vis[maxn];
struct Edge {
int next, to, len;
} edge[maxn << 2];
struct node {
int x, len;
friend bool operator < (node a, node b) {
return a.len > b.len;
}
};
void init() {
memset(edge, 0, sizeof edge);
memset(head, 0, sizeof head);
memset(lay, 0, sizeof lay);
memset(vis, false, sizeof vis);
cnt = 0;
fill(path, path + maxn, INF);
}
void add_edge(int x, int y, int len) {
edge[++cnt].next = head[x];
edge[cnt].to = y;
edge[cnt].len = len;
head[x] = cnt;
}
void spfa() {
queue<int> q;
q.push(1);
path[1] = 0;
while (!q.empty()) {
int now = q.front(); q.pop();
vis[now] = false;
//printf("now : %d %d\n", now, path[now]);
for (int i = head[now]; i; i = edge[i].next) {
int next = edge[i].to, len = edge[i].len;
if (path[next] > path[now] + len) {
path[next] = path[now] + len;
if (!vis[next]) {
q.push(next);
vis[next] = true;
}
}
}
}
}
void dij() {
priority_queue<node> q;
q.push({1, 0});
path[1] = 0;
while (!q.empty()) {
node now = q.top(); q.pop();
if (vis[now.x]) continue;
//printf("now:%d %d\n", now.x, path[now.x]);
vis[now.x] = true;
for (int i = head[now.x]; i; i = edge[i].next) {
int next = edge[i].to, len = edge[i].len;
if (path[next] > path[now.x] + len) {
path[next] = path[now.x] + len;
q.push({next, path[next]});
}
}
}
}
void creat_point(int n, int c) {
add_edge(n + 1, 2*n + 2, c);
for (int i = 2; i < n; i++) {
add_edge(n + i, 2 * n + i - 1, c);
add_edge(n + i, 2 * n + i + 1, c);
}
add_edge(2 * n, 3 * n - 1, c);
}
int main()
{
scanf("%d", &T);
while (T--) {
init();
scanf("%d %d %d", &n, &m, &c);
creat_point(n, c);
for (int i = 1; i <= n; i++) {
scanf("%d", &x);
add_edge(i, x + n, 0);
add_edge(x + 2 * n, i, 0);
}
for (int i = 0; i < m; i++) {
scanf("%d %d %d", &u, &v, &l);
add_edge(u, v, l);
add_edge(v, u, l);
}
//spfa();
dij();
printf("Case #%d: ", ++ca);
if (path[n] != INF) printf("%d\n", path[n]);
else printf("-1\n");
}
return 0;
}