Hdu3666 the matrix problem --- difference Constraint

It would be crazy if I had to get killed from re to TLE To wa ..

This question is not as straightforward as creating a graph. I cannot think of how to create a graph through the given inequality relationship.

Therefore, we need to simplify the conditions given by the question to a certain extent, and take the logarithm of both sides of the inequality to simplify the form of SA-Sb <= C.

Note that W is of the double type.

Second, we will add pruning based on experience:

1. Number of teams> SQRT (n) indicates a negative ring.

2. Count the total number of queues. If the number is greater than 2n, a negative ring is determined.

Generally, this is not used because it is not rigorous.

The two spfa below are both correct. The handwriting queue is a little faster, and the second pruning effect is obvious.

# Include <iostream> # include <cstdio> # include <cstring> # include <cmath> # include <queue> using namespace STD; struct node {int V; double W; int next;} e [360010]; int n, m, H, head [810], INQ [810], outq [810], Q [50000]; double D [805]; void addedge (int A, int B, double C) {e [H]. V = B; E [H]. W = C; E [H]. next = head [a]; head [a] = H ++;} bool spfa (INT s) {int IQ, I, top, K; for (I = 0; I <= N; I ++) d [I] = 1000000000; memset (INQ, 0, sizeof INQ); memse T (outq, 0, sizeof outq); D [s] = 0; INQ [s] = 1; IQ = 0; I = 0; Q [IQ ++] = s; while (I! = IQ) {Top = Q [I]; INQ [Top] = 0; outq [Top] ++; If (outq [Top]> (INT) SQRT (N * 1.0) return 0; k = head [Top]; while (k> = 0) {If (d [E [K]. v]-E [K]. w> d [Top]) {d [E [K]. v] = E [K]. W + d [Top]; If (! INQ [E [K]. v]) {INQ [E [K]. v] = 1; Q [IQ ++] = E [K]. V ;}} K = E [K]. next;} I ++;} return 1;} int spfa (INT st) // adjacent table STL {for (INT I = 0; I <= N; I ++) d [I] = 100000000; memset (INQ, 0, sizeof INQ); memset (outq, 0, sizeof outq); D [st] = 0; INQ [st] = 1; queue <int> q; q. push (ST); int CNT = 1; while (! Q. empty () {int x = Q. front (); q. pop (); INQ [x] = 0; outq [x] ++; If (outq [x]> SQRT (N * 1.0) + 10) return 0; for (INT I = head [X]; I! =-1; I = E [I]. next) {If (d [E [I]. v]> d [x] + E [I]. w) {d [E [I]. v] = d [x] + E [I]. w; If (! INQ [E [I]. v]) {CNT ++; If (CNT> (n + M) * 2) return 0; INQ [E [I]. v] = 1; q. push (E [I]. v) ;}}}return 1 ;}int main () {Double X, ll, UU, L, U; int I, j; while (~ Scanf ("% d % lf", & N, & M, & L, & U) {memset (Head,-1, sizeof head ); H = 0; LL = Log (l); UU = Log (U); for (I = 0; I <n; I ++) for (j = 0; j <m; j ++) {scanf ("% lf", & X); addedge (I, j + N, log (x)-LL ); addedge (J + N, I, UU-log (x);} n + = m; m + = m; If (spfa (0 )) printf ("Yes \ n"); else printf ("NO \ n");} return 0 ;}