Hnu 13411 Reverse a Road II (maximum flow +bfs) Classic

Reverse a Road II

Time limit: 10000ms, special time limit:25000ms, Memory Limit:65536KB

Total Submit users: Accepted users: 6

problem 13411: No Special Judgement

Problem description

JAG Kingdom is a strange kingdom such, it N cities is connected only by one-way roads. The n cities is numbered 1 through N. ICPC (International characteristic Product Corporation) transports its products fro M the factory at the city S to the storehouse on the city T in JAG Kingdom every day. For efficiency, ICPC uses multiple trucks at once. Each truck starts from S and reaches T on the one-way road network, passing through some cities (or directly). In order to reduce risks of traffic jams and accidents, no pair of trucks takes the same road. Now, ICPC wants to improve the efficiency of daily transports, while ICPC operates daily transports by as many trucks as P Ossible under the above constraint. JAG Kingdom, whose finances is massively affected by ICPC, considers to change the direction of one-way roads in order to Increase the number of trucks for daily transports of ICPC. Because reversal of many roads causes confusion, JAG kingdom decides to reverse at more than a single road. If There is no road such that reversal of the road can improve the transport efficiency, JAG Kingdom need not reverse any Roads. Check whether reversal of a single road can improve the current maximum number of trucks for daily transports. And if so, calculate the maximum number of trucks which take disjoint sets of roads when a one-way road can be reversed, a nd the number of roads which can is chosen as the road to being reversed to realize the maximum.

Input

The input consists of multiple datasets. The number of the dataset is no more than 100. Each dataset is formatted as follows. The following M lines describe the information of the roads. The i-th line of them contains and integers aiand bi (1≤ai,bi≤n, ai≠bi), meaning that the I-th road are directed from AI to BI. The end of input is indicated by a line containing four zeros.

Output

For each dataset, output of the integers separated by a single space in a line as follows:if reversal of a single road Impro Ves the current maximum number of trucks for daily transports, the first output integer is the new maximum after reversal Of a road, and the second output integer is the number of roads which can being chosen as the road to being reversed to realize The new maximum. Otherwise, i.e. if the current maximum cannot being increased by any reversal of a road, the first output integer is the Curr ENT maximum and the second output integer is 0.

Sample Input

4 4 1 41 23 14 23 47 8 1 71 21 32 43 44 54 65 77 66 4 5 21 21 34 55 610 21 9 109 19 29 39 410 110 210 310 41 52 52 63 63 7 4 74 81 85 106 107 1010 810 92 15 1 21 21 21 21 21 21 21 21 21 21 21 21 21 21 21 20 0 0 0

Sample Output

1 22 10 04 615 0

Problem Source

JAG practice Contest for ACM-ICPC Asia Regional 2014 Test instructions: Give a direction graph, n points, M edge, beginning S, end point T. Many of the existing robots to go from S to T, each side can only walk a robot, now you can change the direction of a side, ask the maximum number of robots can go from S to T, and how many kinds of modification. (If the direction of the edge is not changed, the method is 0.) ) Problem solving: If you want to not add conditions: You can change a side. So the question is to find out how many ways to go from the beginning to the T. Map: Each Benquan with a forward edge is 1, run one time with the maximum flow. So now to change the direction of a side: if the edge to be changed is traversed by the edge, then will not make the way to increase, there can also make the number of roads reduced, so the edge must not be full flow of the edge (that is, through the edge), now to change a not through the side < U, v > into <v, U> So that the flow can be increased, so long as from s along the not traversed edge to V, and then from u along the side not traversed to T, so that the number of roads can be increased. So we can mark sf[with BFS] from the edge of the residual flow >0 of the maximum flow on each edge of the S-point to some point V, and then Mark Tf[] can go from the T point along the edge of the maximum flow of the residual flow ==0 to some point u (that is, you can walk from the side of the edge to T). Finally, as long as the enumeration is not traversed by the edge of <u, V, if: sf[v]==1&&tf[u]==1 change this side direction can be increased, the number of scenarios k+1. The last maximum number of machines ans = maxflow + k>0? 1:0. #include <stdio.h> #include <string.h> #include <queue> #include <algorithm>using namespace std ; #define Captype intconst int maxn = 1010; Total number of points const int MAXM = 40010; Total number of edges const int INF = 1<<30;struct edg{int to,next; Captype Cap;} Edg[maxm];int Eid,head[maxn];int GAP[MAXN]; The number of points for each distance (or can be considered a height) int DIS[MAXN]; The shortest distance from each point to the end Enode int CUR[MAXN]; Cur[u] indicates that from the U point can flow through the cur[u] number side int pre[maxn];void init () {eid=0; memset (head,-1,sizeof (Head));} There are three parameters to the edge, 4 parameters void addedg (int u,int v,captype c,captype rc=0) without a forward edge {edg[eid].to=v; edg[eid].next=head[u]; Edg[eid].cap=c; head[u]=eid++; Edg[eid].to=u; EDG[EID].NEXT=HEAD[V]; EDG[EID].CAP=RC; head[v]=eid++;} Captype maxflow_sap (int snode,int enode, int n) {//n is the total number of points including the source and sink points, this must be noted memset (Gap,0,sizeof (GAP)); memset (dis,0,sizeof (dis)); memcpy (cur,head,sizeof (head)); Pre[snode] =-1; Gap[0]=n; Captype ans=0; Max Stream int U=snode; while (dis[snode]<n) {//Judging from SNode point there is no flow to the next adjacent point if (U==enode) {//Find an extensible path for (int i=pre[u]; i!=-1; i=pre[edg[i^1].to]) {edg[i].cap-=1; Edg[i^1].cap+=1; Flow of the side that can be recycled} ans+=1; U=snode; Continue } bool flag = FALSE; It is possible to determine whether an int v can flow toward the neighboring point from the U point; for (int i=cur[u]; i!=-1; i=edg[i].next) {v=edg[i].to; if (edg[i].cap>0 && dis[u]==dis[v]+1) {flag=true; Cur[u]=pre[v]=i; Break }} if (flag) {u=v; Continue }//If a flow adjacent point is not found above, then the distance from the starting point U (which can also be considered a height) is the minimum distance of +1 int mind= n for the adjacent flow point; for (int i=head[u]; i!=-1; i=edg[i].next) if (edg[i].cap>0 && mind>dis[edg[i].to]) {Mind=dis [Edg[i].to]; Cur[u]=i; } gap[dis[u]]--; if (gap[dis[u]]==0) return ans; When Dis[u] This distance point is gone, it is impossible to find an augmented stream path from the source point//Because the sink point to the current pointThe distance is only one, then from the source to the meeting point must pass through the current point, but the current point and can not find the point to flow, then the inevitable flow dis[u]=mind+1;//if a stream of adjacent points is found, the distance is the adjacent point distance of +1, if not found, then n+1 gap[dis[u ]]++; if (U!=snode) u=edg[pre[u]^1].to; Return an Edge} return ans; BOOL FLAG[MAXM], SF[MAXN], tf[maxn];void bfs (int s, BOOL flag) {queue<int>q; int u, v; Q.push (s); while (!q.empty ()) {u = Q.front (); Q.pop (); for (int i=head[u]; ~i; i=edg[i].next) {v=edg[i].to; if (flag) {if (sf[v]| |! EDG[I].CAP) continue; Sf[v]=1; Q.push (v); } else{if (tf[v]| |! EDG[I^1].CAP) continue; Tf[v]=1; Q.push (v); }}}}inline void scanf (int& num) {char ch; while (Ch=getchar ()) {if (ch>= ' 0 ' &&ch<= ' 9 ') break; num = ch-' 0 '; while (Ch=getchar ()) {if (ch< ' 0 ' | | Ch> ' 9 ') break; num = num*10+ch-' 0 '; }}int Main () {int n,m, VS, VT, u, v; while (scanf ("%d%d%d%d", &AMP;N,&AMP;M,&AMP;VS,&AMP;VT) >0&&n+m+vs+vt!=0) {init (); for (int i=0; i<m; i++) {scanf (U); scanf (v); ADDEDG (U, V, 1); } memset (flag, 0, (m+3) *sizeof (bool)); memset (sf,0,sizeof (SF)); memset (tf,0,sizeof (TF)); int ans, k=0; Ans = MAXFLOW_SAP (VS, VT, N); for (int i=0; i<m; i++) if (edg[i<<1].cap==0) flag[i]=1; Sf[vs] = 1; BFS (VS, 1); TF[VT] = 1; BFS (VT, 0); for (int i=0; i<m; i++) if (!flag[i]) {v=edg[i<<1|1].to; u=edg[i<<1].to; if (Sf[u]&&tf[v]) k++; } if (k) ans++; printf ("%d%d\n", ans, k); }}

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Hnu 13411 Reverse a Road II (maximum flow +bfs) Classic