"Bzoj" 1631: [Usaco2007 Feb]cow Party (Dijskstra)

http://www.lydsy.com/JudgeOnline/problem.php?id=1631

See m<=100000 decisively with SPFA (but like Dij than SPFA is still slower here?) ）

#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream > #include <algorithm> #include <queue>using namespace std; #define REP (i, n) for (int i=0; i< (n); ++i) # Define FOR1 (I,a,n) for (int i= (a), i<= (n), ++i) #define FOR2 (i,a,n) for (int i= (a);i< (n), ++i) #define FOR3 (I,a,n) for ( int i= (a); i>= (n); i) #define FOR4 (i,a,n) for (int i= (a);i> (n); i) #define CC (i,a) memset (i,a,sizeof (i)) #define Read (a) a=getint () #define PRINT (a) printf ("%d", a) #define DBG (x) cout << #x << "=" << x << endl# Define Printarr (A, N, m) Rep (AAA, N) {rep (BBB, m) cout << a[aaa][bbb]; cout << Endl;} inline const int Getint () {int r=0, k=1; char C=getchar (); for (; c< ' 0 ' | | C> ' 9 '; C=getchar ()) if (c== '-') k=-1; for (; c>= ' 0 ' &&c<= ' 9 '; C=getchar ()) r=r*10+c-' 0 '; return k*r; }inline const int MAX (const int &a, const int &b) {return a>b?a:b;} inline const int min (const int &a, const int &b) {return a<b?a:b;} const int n=1005, m=100005, Oo=~0u>>2;int Ihead[n], N, M, D[n], T, CNT, d1[n], x[m], y[m], w[m];struct ED {int to, Next, W; }e[m];struct nd {int id; const BOOL operator< (const ND &b) const {return d[id]>d[b.id];}}; priority_queue<nd> q;void Add (int u, int v, int w) {e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].w=w;} void Dij (int s) {For1 (i, 0, N) d[i]=oo;d[s]=0; ND t={s};int u, V;q.push (t), while (Q.size ()) {U=q.top (). ID; q.pop (); for (int i=ihead[u]; i; i=e[i].next) if (d[v=e[i].to]& GT;D[U]+E[I].W) {D[v]=d[u]+e[i].w;t.id=v;q.push (t);}}} int main () {read (n); read (m); Read (T); int u, V, w;rep (i, m) {read (u); Read (v); Read (w); X[i]=u; Y[i]=v; W[i]=w;add (V, U, W);} int Mx=0;dij (T); For1 (i, 1, n) d1[i]=d[i];cnt=0; CC (ihead, 0); Rep (I, M) add (X[i], y[i], w[i]);d ij (T); For1 (i, 1, n) {mx=max (MX, d1[i]+d[i]);} Print (MX); return 0;}

Description Farm has N (1≤n≤1000) cattle stalls, each of which has 1 cows to participate in the X Cow cattle party. A total of M (1≤m≤100000) A one-way link to the barn, article I crippled need ti time to pass. Cows are lazy, so whether they go to the X-Barn party or return to their home, they use the least-used route. So how much time does it take to go back and forth with the biggest cows? Input

Line 1th: Three integers separated by a space.

Line 2nd to line m+1, three integers separated by spaces: Ai, Bi, and Ti. Represents the starting point of a road, the end point, and the time it takes.

Output

Unique line: An integer: The maximum amount of time that will be spent in all cows participating in the party.

Sample INPUT4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3
Sample Output10
HINT

Sample Description:

There were 4 cows at the party, 8 Roads and a party on the 2nd farm.

The 4th cow can go directly to the party location (spending 3 time), then the return route passes through the 1th and 3rd farms (7 time), for a total of 10 hours.

Source

Silver

"Bzoj" 1631: [Usaco2007 Feb]cow Party (Dijskstra)