HDU1535 Invitation Cards Shortest circuit

Portal: "HDU" 1535 invitation Cards

Topic Analysis: The topic is really difficult to read ... In fact, the problem is to ask for the sum of the shortest distance from the point numbered 1 to all other points, plus all other points to the shortest distance of the points numbered 1. As long as the original image and the reverse diagram to run the shortest time is good.

The code is as follows:

#include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define REP (I, A, b) for (int i = (a); I < (b); + + i) #define for (I, A, b) for (int i = (a); I < = (b); + + i) #define REV (I, A, b) for (int i = (a); I >= (b);---i) #define Travel (E, H, u) for (edge* e = h [u]; E
E = e = next) #define CLR (A, X) memset (A, x, sizeof a) const int MAXN = 1000005;
const int MAXH = 2000005;
const int maxe = 2000005;

const int INF = 0X3F3F3F3F;
	struct Edge {int V, C;
Edge* Next;

} ;
	
	struct Heap {int V, IDX; Heap () {} heap (int v, int idx): V (v), IDX (IDX) {} BOOL operator < (const heap& a) const {RE
	Turn v < A.V;

}
} ;
	struct Priority_queue {Heap heap[maxh];
	
	int point;
	Priority_queue (): Point (1) {} void Clear () {point = 1;
	} bool Empty () {return point = = 1;
		} void maintain (int o) {int x = O; while (o &GT
			1 && heap[o] < heap[o >> 1]) {swap (Heap[o], heap[o >> 1]);
		o >>= 1;
		} o = x; int p = o, L = o << 1, r = O << 1 |
		1;
			while (O < point) {if (L < point && Heap[l] < heap[p]) P = l;
			if (R < point && Heap[r] < heap[p]) P = r;
			if (p = = O) break;
			Swap (Heap[o], heap[p]); o = p, l = o << 1, r = O << 1 |
		1;
		}} void push (int v, int idx) {Heap[point] = heap (v, IDX);
	Maintain (point + +);
		} void Pop () {heap[1] = heap[--Point];
	Maintain (1);
	} int Front () {return heap[1].idx;
	} Heap Top () {return heap[1];

}
} ;
	struct Shortest_path_algorithm {priority_queue q;
	Edge E[maxe], *H[MAXN], *cur;
	Edge Re[maxe], *RH[MAXN], *rcur;
	int D[MAXN];
	BOOL VIS[MAXN];
	
	int Q[MAXN], head, tail;
		void init () {cur = E;
		Rcur = RE;
		CLR (H, 0);
	CLR (RH, 0); } void Addedge (intu, int v, int c) {cur v = v;
		Cur c = c;
		Cur next = h[u];
		H[u] = cur + +;
		Rcur v = u;
		Rcur C = C;
		Rcur next = Rh[v];
	RH[V] = rcur + +;
		} void Dijkstra (int s, edge* h[]) {q.clear ();
		CLR (d, INF);
		CLR (Vis, 0);
		D[s] = 0;
		Q.push (D[s], s);
			while (!q.empty ()) {int u = q.front ();
			Q.pop ();
			if (Vis[u]) continue;
			Vis[u] = 1;
				Travel (E, H, u) {int v = e, V, C = e-C;
					if (D[v] > D[u] + c) {D[v] = D[u] + C;
				Q.push (D[v], V);
		}}}} void Spfa (int s, edge* h[]) {head = tail = 0;
		CLR (d, INF);
		CLR (Vis, 0);
		D[s] = 0;
		Q[tail + +] = s;
			while (head! = tail) {int u = q[head + +];
			if (head = = MAXN) head = 0;
			Vis[u] = 0;
				Travel (E, H, u) {int v = e, V, C = e-C;
					if (D[v] > D[u] + c) {D[v] = D[u] + C;
if (!vis[v]) {vis[v] = 1;						if (D[v] < D[q[head]]) {if (head = = 0) head = MAXN;
						q[--Head] = v;
							} else {Q[tail + +] = V;
						if (tail = = maxn) tail = 0;

}}}}}} G;


int n, m;
	void scanf (int& x, char c = 0) {while ((c = GetChar ()) < ' 0 ' | | c > ' 9 ');
	x = C-' 0 ';
while ((c = GetChar ()) >= ' 0 ' && C <= ' 9 ') x = x * + C-' 0 ';
	} void Solve () {int u, V, c;
	int ans = 0;
	G.init ();
	SCANF (n), scanf (m);
		while (M--) {scanf (U), scanf (v), scanf (c);
	G.addedge (U, V, c);
	} g.dijkstra (1, G.H);
	For (i, 1, n) ans + = g.d[i];
	G.dijkstra (1, G.RH);
	For (i, 1, n) ans + = g.d[i];
printf ("%d\n", ans);
	} int main () {int T;
	scanf ("%d", &t);
	while (T--) solve ();
return 0; }