HDU 3416 marriage match IV (spfa + maximum Stream)

The general meaning of the question is: give you some directed edge for you to find the given point S, the number of the shortest circuit between T.

The spfa on both sides are from S to T, and from t to S, and then the vertex on the shortest road is obtained to build an edge with a capacity of 1, and then the maximum traffic from S to T is obtained, is the number of the most short circuits.

PS: the code writing posture is not beautiful enough.

Marriage match IV Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 2051 accepted submission (s): 608


Problem descriptiondo not sincere non-interference.
Like that show, now starvae also take part in a show, but it take place between City A and B. starvae is in City A and girls are in City B. every time starvae can get to City B and make a data with a girl he likes. but there are two problems with it, one is starvae must get to B within least time, it's said that he must take a shortest path. other is no road can be taken more than once. while the city starvae passed away can been taken more than once.


So, under a good RP, starvae may have chances to get to City B. but he don't know how many chances at most he can make a data with the girl he likes. cocould you help starvae?
Inputthe first line is an integer t indicating the case number. (1 <= T <= 65)
For each case, there are two integer N and m in the first line (2 <= n <= 1000, 0 <= m <= 100000 ), n is the number of the City and m is the number of the roads.

Then follows M line, each line have three integers A, B, C, (1 <= A, B <= N, 0 <C <= 1000) it means there is a road from A to B and it's distance is C, while there may have no road from B to. there may have a road from A to A, but you can ignore it. if there are two roads from A to B, they are different.

At last is a line with two integer a and B (1 <= A, B <= n,! = B), means the number of city a and City B.
There may be some blank line between each case.
Outputoutput a line with a integer, means the chances starvae can get at most.
Sample Input

37 81 2 11 3 12 4 13 4 14 5 14 6 15 7 16 7 11 76 71 2 12 3 11 3 33 4 13 5 14 6 15 6 11 62 21 2 11 2 21 2

 
Sample output

211

# Include <algorithm> # include <iostream> # include <stdlib. h> # include <string. h> # include <iomanip> # include <stdio. h> # include <string> # include <queue> # include <cmath> # include <stack> # include <map> # include <set> # define EPS 1e-12 ///# define M 1000100 # define ll _ int64 // # define ll long // # define INF 0x7ffffff # define INF 0x3ffffff # define PI 3.1415926535898 # define zero (X) (FABS (x) <EPS )? 0: X) using namespace STD; const int maxn = 2010; int CNT; int n, m; int cur [maxn], head [maxn]; int dis [maxn], gap [maxn]; int Aug [maxn], pre [maxn]; struct node {int V, W; int next;} f [510000]; int head1 [maxn]; int head2 [maxn]; int cnt1; int cnt2; struct node1 {int U, V, W; int next ;}; node1 PP [500010], FF [500010]; void Init () {CNT = 0; cnt1 = 0; cnt2 = 0; memset (head1,-1, sizeof (head1); memset (Head,-1, sizeof (H EAD); memset (head2,-1, sizeof (head2);} void Add1 (int u, int V, int W) {PP [cnt1]. U = u; PP [cnt1]. V = V; PP [cnt1]. W = W; PP [cnt1]. next = head1 [u]; head1 [u] = cnt1 ++;} void Add2 (int u, int V, int W) {FF [cnt2]. U = u; FF [cnt2]. V = V; FF [cnt2]. W = W; FF [cnt2]. next = head2 [u]; head2 [u] = cnt2 ++;} void add (int u, int V, int W) {f [CNT]. V = V; F [CNT]. W = W; F [CNT]. next = head [u]; head [u] = CNT ++; F [CNT]. V = U; F [CNT]. W = 0; F [CNT]. next = head [v]; head [v] = CNT ++;} int SAP (int s, int e, int N) {int max_flow = 0, V, U = s; int ID, mindis; Aug [s] = inf; Pre [s] =-1; memset (DIS, 0, sizeof (DIS); memset (gap, 0, sizeof (GAP); Gap [0] = N; For (INT I = 0; I <= N; ++ I) cur [I] = head [I]; // initialize the current arc as the first arc while (DIS [s] <n) {bool flag = false; if (u = e) {max_flow + = Aug [E]; for (V = pre [E]; V! =-1; V = pre [v]) // path tracing updates residual network {id = cur [v]; F [ID]. w-= Aug [E]; F [ID ^ 1]. W + = Aug [E]; Aug [v]-= Aug [E]; // modify the augmented quantity. If (F [ID] will be used in the future. W = 0) u = V; // do not roll back to the source point, only roll back to the arc end of the arc with the capacity of 0} For (ID = cur [u]; ID! =-1; id = f [ID]. next) // start from the current arc to find the allowable arc {v = f [ID]. v; If (F [ID]. w> 0 & dis [u] = dis [v] + 1) // find the allowable arc {flag = true; Pre [v] = u; cur [u] = ID; Aug [v] = min (Aug [u], F [ID]. w); U = V; break;} If (flag = false) {If (-- gap [dis [u] = 0) break; /// gap optimization. When a fault occurs in the hierarchy tree, the end algorithm mindis = N; cur [u] = head [u]; for (ID = head [u]; ID! =-1; id = f [ID]. next) {v = f [ID]. v; If (F [ID]. w> 0 & dis [v] <mindis) {mindis = dis [v]; cur [u] = ID; /// modify the current arc while modifying the label} dis [u] = mindis + 1; Gap [dis [u] ++; If (u! = S) u = pre [u]; // trace back to continue searching for allowed arcs} return max_flow;} int D1 [maxn]; int vis [maxn]; queue <int> FP; void spfa1 (INT s) {for (INT I = 0; I <= N; I ++) D1 [I] = inf; while (! FP. empty () FP. pop (); memset (VIS, 0, sizeof (VIS); vis [s] = 1; FP. push (s); D1 [s] = 0; while (! FP. empty () {int x = FP. front (); vis [x] = 0; FP. pop (); For (INT I = head1 [X]; I! =-1; I = PP [I]. next) {int v = PP [I]. v; If (d1 [v]> d1 [x] + PP [I]. w) {D1 [v] = D1 [x] + PP [I]. w; If (! Vis [v]) {FP. push (V); vis [v] = 1 ;}}}} int D2 [maxn]; void spfa2 (INT s) {for (INT I = 0; I <= N; I ++) D2 [I] = inf; while (! FP. empty () FP. pop (); memset (VIS, 0, sizeof (VIS); vis [s] = 1; FP. push (s); d2 [s] = 0; while (! FP. empty () {int x = FP. front (); FP. pop (); vis [x] = 0; For (INT I = head2 [X]; I! =-1; I = FF [I]. next) {int v = FF [I]. v; If (D2 [v]> D2 [x] + FF [I]. w) {d2 [v] = d2 [x] + FF [I]. w; If (! Vis [v]) {FP. push (V); vis [v] = 1 ;}}} int main () {int K; CIN> K; while (k --) {Init (); scanf ("% d", & N, & M); int X, Y; int U, V, W; For (INT I = 0; I <m; I ++) {scanf ("% d", & U, & V, & W); Add1 (U, V, W); Add2 (V, u, W) ;}scanf ("% d", & X, & Y); spfa1 (x); spfa2 (y); For (INT I = 0; I <cnt1; I ++) if (d1 [PP [I]. u] + D2 [PP [I]. v] + PP [I]. W = D1 [y]) add (PP [I]. u, PP [I]. v, 1); int ans = SAP (X, Y, n); cout <ans <Endl;} return 0 ;}