ZOJ2770, fire connection, differential Constraint

The key to difference constraints is to understand why they can be converted into triangular inequalities.

In addition, for the inequality Xi-Xj <= c, it must be converted to an edge. .

The ZOJ2770 analysis is correct.

The specific analysis is not written, and the code = may be slightly messy.

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           Using namespace std; # define clr (x) memset (x, 0, sizeof (x) # define fp1 freopen ("in.txt", "r", stdin) # define fp2 freopen ("out.txt", "w", stdout) # define pb push_back # define INF 0x3c3c3c3ctypedef long LL; struct Edge {LL from, to, dist ;}; const int maxn = 1e6; struct BF {// The node number should start from 0. It should be possible not to start from 0 int n; vector
          
            Edges; vector
           
             G [maxn]; bool inq [maxn]; // queue optimization spfa determines whether edges have been added to LL d [maxn]; int p [maxn]; // The Last arc int is_neg [maxn] in the shortest path; // The number of times the queue enters is used to determine whether there is a negative weight loop void init () {for (int I = 0; I <= n; I ++) G [I]. clear (); edges. clear ();} void addedge (LL from, LL to, LL dist) // unidirectional edge operation {// printf ("% d ~ \ N ", from, to, dist); edges. push_back (Edge) {from, to, dist}); int len = edges. size (); G [from]. push_back (len-1);} int spfa (LL be) // bool can be returned to determine the negative weight ring {memset (inq, 0, sizeof (inq); memset (p, 0, sizeof (p); memset (is_neg, 0, sizeof (is_neg); // The comment is used to judge the negative ring queue.
            
              Q; for (int I = 0; I <= n; I ++) d [I] = INF; d [be] = 0; Q. push (be); inq [be] = true; while (! Q. empty () // non-empty ~ {Int u = Q. front (); Q. pop (); inq [u] = false; for (int I = 0; I <G [u]. size (); I ++) {Edge & e = edges [G [u] [I]; if (d [e. to]> d [u] + e. dist) {// the value you want to input has been converted to e. dist d [e. to] = d [u] + e. dist; p [e. to] = u; // G [u] [I] stores the information of an edge, which does not exist and is directly changed to u if (! Inq [e. to]) {Q. push (e. to); inq [e. to] = true; if (++ is_neg [e. to]> n) return false;} // else ...}} return true ;}} test; LL sold [1000 + 10]; LL sum [1000 + 10]; int main () {// fp1; LL n, m,, b, c; while (scanf ("% lld", & n, & m) = 2) {test. n = n; test. init (); clr (sold); clr (sum); for (int I = 1; I <= n; I ++) {scanf ("% lld ", & sold [I]); sum [I] = sum [I-1] + sold [I]; test. addedge (I, I-1, 0); // Si-Si-1> 0 per barracks Cannot be less than 0 test. addedge (I-1, I, sold [I]); // up to sold [I];} // puts ("-----------------"); for (int I = 1; I <= m; I ++) {scanf ("% lld", & a, & B, & c); test. addedge (B, A-1,-c); // a --- B at least c People // test. addedge (A-1, B, sum [B]-sum [A-1]); // a --- B up to x people} if (! Test. spfa (n) {puts ("Bad Estimations"); // printf ("% d \ n", test. d [0]);} else printf ("% d \ n",-test. d [0]); // test. n = 3; // test. addedge (1, 2, 5); // test. addedge (2, 3, 6); // if (! Test. spfa (1) puts ("hehe"); // else printf ("% d \ n", test. d [1], test. d [2], test. d [3]);} return 0 ;}