HDU 5406 CRB and Apple pattern cost flow Solution

Test instructions Simplification is a sequence that finds two of the largest non-descending sub-sequences making them lengths and longest.

= = as a DP slag, the state design is probably to the dp[i][j] means the first person to take the last I, the second person took to j this place. How to transfer within the feasible complexity? No, look at the sky.

In fact, as a graph theory workers first reaction is the cost of flow, but the number of sides too many do not dare to engage in =

However, the cost flow can also be too. (x, Y) indicates that the capacity is X cost for Y,

It is simple to build a source s and sink E, plus a point t limit the maximum flow from the source is 2, which is s->t (2, 0). Remove each point can only be taken once, x->x ' (1, 1), capacity 1 limit can only be taken once, the cost represents the length of +1

T->x (1, 0) can start at any point. X '->e (1, 0) can end at any one point.

Run the maximum cost of the maximum flow and then it will be T (hello)

Remember POJ a problem card SPFA, but change into a stack can be over = =

So you can change it to a stack and find that it's not T. (From the outset, this problem has been bothering me. T-t How to construct the card queue not stack data? T-T for guidance)

Code 1, about 1400ms.

#include <cstdio>#include<cstring>#include<algorithm>#include<queue>#include<stack>using namespacestd;Const intN = .;Const intM =4000005;Const intINF =0x3f3f3f3f;structpoint{intu,v, Next, flow, cost;    Point () {}; Point (intXintYintZintFintc) {u= x, v= y; Next = Z, flow = f, cost =C; };} P[M];intPre[n], head[n], d[n];BOOLVis[n];intS, E, no;structinfo{intH, D; voidScan () {scanf ("%d%d", &h, &d); }    BOOL operator< (ConstInfo &i)Const{        if(h = = I.h)returnD <I.D; returnH >I.h; }}info[n];BOOLSPFA () {stack<int>Q; intx, Y, now, I; memset (d,0xc0,sizeof(d)); memset (Vis,0,sizeof(VIS)); memset (PRE,-1,sizeof(pre));  Q.push (s); D[s]=0; Vis[s] =1;  while(!Q.empty ()) {x=Q.top ();        Q.pop (); VIS[X]=0;  for(i = head[x]; I! =-1; i =P[i].next) {y=p[i].v; if(P[i].flow && D[y] < d[x] +p[i].cost) {D[y]= D[x] +P[i].cost; Pre[y]=i; if(!Vis[y])                    {Q.push (y); Vis[y]=1; }            }        }    }    return(D[e]! = d[e+1]);}intMCMF () {intMaxflow =0, I, minflow, mincost =0;  while(SPFA ()) {Minflow= INF +1;  for(i = pre[e]; I! =-1; i =PRE[P[I].U]) {            if(P[i].flow <minflow) Minflow=P[i].flow; }         for(i = pre[e]; I! =-1; i =PRE[P[I].U]) {P[i].flow-=Minflow; P[i^1].flow + =Minflow; } mincost+ = d[e] *Minflow; }    returnMincost;}voidinit () {memset (head,-1,sizeof(head)); No=0;}voidAddintXintYintFintCO) {P[no]= Point (x, Y, head[x], F, CO); HEAD[X] = no++; P[no]= Point (Y, X, Head[y],0,-co); Head[y] = no++;}intMain () {intTC, N, I, J; scanf ("%d", &TC);  while(tc--) {scanf ("%d", &N); Init (); s=0; E =2* n +2;  for(i =1; I <= N; i++) {Info[i].scan (); Add (i<<1, I <<1|1,1,1); Add (1, I <<1,1,0); Add (i<<1|1E1,0); } Sort (Info+1, info + n +1);  for(i =1; I <= N; i++){             for(j = i +1; J <= N; J + +){                if(INFO[I].D <=info[j].d) {Add (i<<1|1, J <<1,1,0); }}} Add (S,1,2,0); printf ("%d\n", MCMF ()); }    return 0;}

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Then proceed to the various optimization stages.

The first attempt to improve the way to build the edge, specifically see http://blog.csdn.net/mxymxy1994mxy/article/details/47818397 the great God, the space saved a lot, However, in the ordinary queue or T (of course, it may be my stupid (but obviously not possible, in fact, is that I am stupid, lying (why do you want to vomit))), simple form is a strong orz, but to the stack of SPFA, less than 500ms, is bad ah.

Code 2, the others are the same, modified the way the edge is built

Init (); s =0; E =2* n +2;  for(i =1; I <= N; i++) {Info[i].scan (); Add (i<<1, I <<1|1,1, -1); Add (1, I <<1,1,0); Add (i<<1|1E1,0); } Sort (Info+1, info + n +1); intMAXN;  for(i =1; I <= N; i++) {MAXN=INF;  for(j = i +1; J <= N; J + +){                if(Info[j].d <info[i].d) {                    Continue; }                if(INFO[J].D <=MAXN) {Add (i<<1|1|1, J <<1,1,0); Add (i<<1, J <<1,1,0); //flag = false;MAXN =INFO[J].D; }}} Add (S,1,2,0);

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Get out and eat dinner and go back to the lab and try all the solutions.

HDU 5406 CRB and Apple pattern cost flow Solution