Poj 3159 candies difference Constraint

Set a [I] to the number of sweets for the I-th child, and D [I] to the number of sweets for the I-th child relative to the 1st child, that is:

D [I] = A [I]-A [1], d [1] = 0.

A B c includes a [B]-A [a] <= C (A [B]-A [1]). -(A [a]-A [1]) <= c, d [B]-d [a] <= C

The question is to find the maximum value, that is, to find the shortest path with 1 as the source point.

Note: The use of spfa + queues will be Tlm. It takes 700 ms to switch to stack. I tried Dijkstra + priority_queue and wiped it over 1400 + Ms. I wonder if it is not well written.

 

# Include <iostream> # include <queue> # include <stack> using namespace STD; const int max = 30005; const int INF = 1000000000; const int n = 150005; struct node {int V; int cost; int next ;}; node [N]; int d [Max]; int adj [Max]; bool in_q [Max]; int CNT [Max]; int size; int n, m; void add_edge (int u, int V, int cost) {node [size]. V = V; node [size]. cost = cost; node [size]. next = adj [u]; adj [u] = size ++;} struct CMP {bool ope Rator () (const Int & A, const Int & B) {return d [a]> d [B] ;}}; priority_queue <int, vector <int>, CMP> q; void Dijkstra () {for (INT I = 0; I <= N; I ++) d [I] = inf; q. push (1); D [1] = 0; int U, V, W; while (! Q. Empty () {u = Q. Top (); q. Pop (); For (INT I = adj [u]; I! =-1; I = node [I]. next) {v = node [I]. v; W = node [I]. cost; If (d [v]> d [u] + W) {d [v] = d [u] + W; q. push (v) ;}}} void spfa () {memset (in_q, false, sizeof (in_q); For (INT I = 1; I <= N; I ++) d [I] = inf; stack <int> S; d [1] = 0; S. push (1); in_q [1] = true; int U, V, W; while (! S. Empty () {u = S. Top (); S. Pop (); in_q [u] = false; For (INT I = adj [u]; I! =-1; I = node [I]. next) {v = node [I]. v; W = node [I]. cost; If (d [v]> d [u] + W) {d [v] = d [u] + W; If (! In_q [v]) {in_q [v] = true; S. push (v) ;}}}} int main () {int A, B, W; scanf ("% d", & N, & M ); for (INT I = 0; I <= N; I ++) adj [I] =-1; for (INT I = 0; I <m; I ++) {scanf ("% d", & A, & B, & W); add_edge (a, B, W);} Dijkstra (); printf ("% d \ n", d [N]); Return 0 ;}