Rogu P2939 [usaco 09feb] rebuild Revamping Trails, p2939revamping

Rogu P2939 [usaco 09feb] rebuild Revamping Trails, p2939revamping
Description

Farmer John dutifully checks on the cows every day. he traverses some of the M (1 <= M <= 50,000) trails conveniently numbered 1 .. M from pasture 1 all the way out to pasture N (a journey which is always possible for trail maps given in the test data ). the N (1 <= N <= 10,000) pastures conveniently numbered 1 .. N on Farmer John's farm are currently connected by bidirectional dirt trails. each trail I connects pastures P1_ I and P2_ I (1 <= P1_ I <= N; 1 <= P2_ I <= N) and requires T_ I (1 <= T_ I <= 1,000,000) units of time to traverse.

He wants to revamp some of the trails on his farm to save time on his long journey. specifically, he will choose K (1 <= K <= 20) trails to turn into highways, which will adjust tively reduce the trail's traversal time to 0. help FJ decide which trails to revamp to minimize the resulting time of getting from pasture 1 to N.

Time limit: 2 seconds

John has N pastures in total. The trails are connected by M Dusty Trails. The trails are accessible in two directions. John leaves from farm 1 every morning to farm N to check his health for the cows.

It takes some time to pass each trail. john planned to upgrade K of the trails to make them high-speed roads. the highway traffic is almost completed in an instant, so the highway traffic time is 0.

Please help john decide which trails to upgrade, so that he spends the least time to reach the farm every morning. output the minimum time.

Input/Output Format

Input Format:

 

* Line 1: Three space-separated integers: N, M, and K

* Lines 2.. M + 1: Line I + 1 describes trail I with three space-separated integers: P1_ I, P2_ I, and T_ I

 

Output Format:

 

* Line 1: The length of the shortest path after revamping no more than K edges

 

Input and Output sample input sample #1: Copy

4 4 1 1 2 10 2 4 10 1 3 1 3 4 100 

Output example #1: Copy

1 

Description

K is 1; revamp trail 3-> 4 to take time 0 instead of 100. The new shortest path is 1-> 3-> 4, total traversal time now 1.

 

If you know what a layered graph is, it's a raw question.

Otherwise, it is a question.

First, it is definitely not easy to process the source image.

Then we layer the graph. Specifically, we create K + 1 source image. The Edge Weight between the graph and the graph is 0, and the edge is the edge in the source image.

In this way, you can run a two-dimensional dijstra (hard to write ).

 

# Include <cstdio> # include <algorithm> # include <queue> # include <cstring> # define Pair pair <int, int> # define F first # define S secondconst int MAXN = 1e6 + 10; using namespace std; inline int read () {char c = getchar (); int x = 0, f = 1; while (c <'0' | c> '9') {if (c = '-') f =-1; c = getchar () ;}while (c >='0' & c <= '9') {x = x * 10 + c-'0 '; c = getchar ();} return x * f;} int N, M, K; int dis [MAXN] [21], vis [MAXN] [21]; struct node {int u, v, w, nxt;} edge [M AXN]; int head [MAXN], num = 1; inline void AddEdge (int x, int y, int z) {edge [num]. u = x; edge [num]. v = y; edge [num]. w = z; edge [num]. nxt = head [x]; head [x] = num ++;} void Dijstra () {memset (dis, 0xf, sizeof (dis )); dis [1] [0] = 0; // vertex I, layer j priority_queue <pair <int, Pair> q; q. push (make_pair (0, make_pair (); // The first vertex represents the second presentation layer while (q. size ()! = 0) {while (vis [q. top (). s. f] [q. top (). s.S] & q. size ()> 0) q. pop (); Pair p = q. top (). second; vis [p. f] [p.S] = 1; for (int I = head [p. f]; I! =-1; I = edge [I]. nxt) {int will = edge [I]. v; if (vis [will] [p.S] = 0 & dis [will] [p.S]> dis [p. f] [p.S] + edge [I]. w) dis [will] [p.S] = dis [p. f] [p.S] + edge [I]. w, q. push (make_pair (-dis [will] [p.S], make_pair (will, p.S ))); if (p.S + 1 <= K & vis [will] [p.S + 1] = 0 & dis [will] [p.S + 1]> dis [p. f] [p.S]) dis [will] [p.S + 1] = dis [p. f] [p.S], q. push (make_pair (-dis [will] [p.S + 1], make_pair (will, p.S + 1) ;}} printf ("% d ", dis [N] [K]);} int main () {memset (head,-1, sizeof (head); N = read (); M = read (); K = read (); for (int I = 1; I <= M; I ++) {int x = read (), y = read (), z = read (); AddEdge (x, y, z); AddEdge (y, x, z);} Dijstra (); return 0 ;}