1062 expensive dowry (spfa + level enumeration)-from lanshui_Yang

Solution: first build the graph, note that it is a directed graph, and then use 1 as the source point, the enumeration level limit, that is to say, each time we use spfa to get 1 point to other points that can be reached (due to the level limit, not all points may be reached in one spfa), and finally find the minimum cost. Ps: the level of point 1 in this question may not be the highest. See the Code:

# Include <iostream> # include <string> # include <algorithm> # include <cstring> # include <queue> # include <cmath> # include <vector> # include <cstdio> # define mem (, b) memset (a, B, sizeof (a) using namespace std; inline void RD (int & a) {a = 0; char t; do {t = getchar () ;}while (t <'0' | t> '9'); a = t-'0 '; while (t = getchar ()> = '0' & t <= '9') {a = a * 10 + t-'0 ';}} inline void OT (Int a) {if (a >=10) {OT (a/10);} putchar (a % 10 + '0');} const int MAXN = 105; const int INF = 0x7fffffff; int m, n; struct Node {int adj; int c ;}; vector <Node> vert [MAXN]; int P [MAXN]; // The price of each item int L [MAXN]; // The level of each item holder int d [MAXN]; // record the Shortest Path void clr () from 1 to other points each time. // initialize {mem (P, 0); mem (L,-1); int I; for (I = 1; I <= n; I ++) vert [I]. clear ();} void init () {clr (); Int I, j; for (I = 1; I <= n; I ++) {scanf ("% d", & P [I], & L [I]); int t; scanf ("% d", & t); for (j = 1; j <= t; j ++) // create a graph {int B; scanf ("% d", & B); Node tmp; tmp. adj = B; scanf ("% d", & tmp. c); vert [I]. push_back (tmp) ;}} bool inq [MAXN]; queue <int> q; void spfa (int M) {while (! Q. empty () q. pop (); q. push (1); d [1] = 0; inq [1] = true; int u; while (! Q. empty () {u = q. front (); q. pop (); inq [u] = false; int I; for (I = 0; I <vert [u]. size (); I ++) {int v = vert [u] [I]. adj; int ci = vert [u] [I]. c; if (L [v]> = M & L [v] <= M + m) {if (d [u]! = INF & d [u] + ci <d [v]) {d [v] = d [u] + ci; if (! Inq [v]) {q. push (v); inq [v] = true ;}}}} void solve () {int X = max (L [1]-m, 0 ); int MIN = P [1]; int M; for (M = X; M <= L [1]; M ++) // enumeration level {int I; for (I = 1; I <= n; I ++) {d [I] = INF;} mem (inq, 0); spfa (M ); for (I = 2; I <= n; I ++) {if (L [I]> = M & L [I] <= M + m & d [I]! = INF & MIN> d [I] + P [I]) // note the judgment condition MIN = d [I] + P [I] ;}} printf ("% d \ n", MIN);} int main () {while (scanf ("% d", & m, & n )! = EOF) {init (); solve ();} return 0 ;}