The game problem of codevs1033 earthworms

Description

On a trapezoidal field, a group of earthworms are doing a collection of food games. The earthworms piled up the food on the trapezoidal field as follows:

A (a) a (1,m)

A (2,1) a (2,2) a (2,3) ... a (2,m) a (2,m+1)

A (3,1) a (3,2) a (3,3) ... a (3,m+1) a (3,m+2)

......

A (n,1) a (n,2) a (n,3) ... a (n,m+n-1)

They divide the food into n rows, the 1th row is the food of the M heap, and the amount of food per heap is a (a), A (,..., a (1,m);

The 2nd row has m+1 heap food, each pile of food quantity is a (2,1), A (2,2),..., A (2,m+1), the following successively M+2 heap, m+3 heap 、... m+n-1 heap food.

Now earthworms have chosen K-worms to test their ability to cooperate (1≤K≤M). The test method is as follows: 1th earthworms choose a pile of food from line 1th, then crawl to the lower left or right, and collect 1 piles of food, for example, from a (2,2) or a (2,3), and not climb to other places. Then crawl down one line and collect a bunch of food until the nth row collects a bunch of food. 1th The amount of food collected by earthworms is the sum of the amount of food it collects in each row; the 2nd worm also climbs from line 1th to Nth row, each row collects a pile of food, crawls the method is similar with the 1th worm, but cannot meet the 1th worm to climb the trajectory; In general, the first earthworm crawls from the 1th line to the nth row, Collect a pile of food per row, the method of crawling is similar to the 1th earthworm, but can not meet the previous I-1 of the worm crawling trajectory. How should this K-worm work together to make the largest amount of food they collect? The amount of food collected can represent the level of cooperation of this K-worm.

• Ø Programming tasks:

Given the trapezoidal m, n, and K values (1≤K≤M≤30;1≤N≤30) and the amount of food per pile in the ladder (non-integers less than 10), the maximum amount of food that can be collected by this K-worm is programmed.

Input Description

The input data is provided by a text file named Input1.*, with a total of n+1 lines. Each row is separated by a space between two data.

The 1th line is the values of N, M, and K.

The next n rows are the trapezoid of each row of food volume A (i,1), A (i,2), ..., a (i,m+i-1), i=1,2,..., N.

Output Description

At the end of the program, the maximum amount of food that can be collected by the K earthworm Bar is output on the screen.

Sample Input

3 2 2

1 2

5 0 2

1 10 0 6

Sample Output

26

Do the codevs1227 box to take the number 2 and then do this problem is very simple. The same modeling method, the same cost flow. For more information, please see Codevs 1227-square Check number 2

Why do we have to split a point into two? In fact, only then can we control a point only to be counted as a contribution in order to ensure that the answer is correct.

The code is as follows:

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5#include <cmath>6 #defineFile (s) freopen (S ".", "R", stdin), Freopen (S ". Out", "w", stdout)7 #defineMAXN 1000108 #defineINF 21474836479 #defineR (J) (J^1)Ten  One using namespacestd; AtypedefLong LongLlg; -  - intN,M,K,JIA,S,T,ANS,L,R,DIS[MAXN],FF[MAXN],FA[MAXN],D1[MAXN],D[MAXN]; the inthead[maxn],next[maxn],to[maxn],f[maxn],c[maxn],tt=1, s1,t1; - BOOLW[MAXN]; -  - voidLinkintXintYintZinto) { +to[++tt]=y;next[tt]=head[x];head[x]=tt; -to[++tt]=x;next[tt]=head[y];head[y]=tt; +c[tt-1]=z; f[tt-1]=o; f[tt]=-o; A } at  - BOOLSPFA () { -      for(intI=1; i<=t;i++) dis[i]=-inf,d1[i]=INF; -L=r=0; D[r++]=s; dis[s]=0; d1[s]=INF; -      while(l!=R) { -         intU=d[l++]; w[u]=0; in         if(L&GT;=MAXN) l=0; -          for(intI=head[u],v;v=to[i],i;i=Next[i]) to             if(c[i]>0&& dis[v]<dis[u]+F[i]) { +dis[v]=dis[u]+F[i]; -Ff[v]=i; fa[v]=u; thed1[v]=min (d1[u],c[i]); *                 if(!W[v]) { $w[v]=1;d [r++]=v;Panax Notoginseng                     if(R&GT;=MAXN) r=0; -                 } the             } +     } A     if(Dis[t]==-inf)return 0; theans+=dis[t]*D1[t]; +      for(intnow=t;now!=s;now=Fa[now]) { -c[ff[now]]-=D1[t]; $C[r (Ff[now])]+=D1[t]; $     } -     return 1; - } the  - intMain () {WuyiFile ("a"); thescanf" %d%d%d",&n,&m,&k); -Jia= (m*2+n-1) *n/2; Wus1=jia*2+1; t1=s1+1; s=t1+1; t=s+1; -Link (s,s1,k,0); Link (t1,t,k,0); About      for(intI=1, now=1; i<=n;i++) $          for(intj=1, x;j<m+i;j++,now++){ -scanf"%d",&x); -             if(i==1) Link (s1,now,1,0); -Link (Now,now+jia,1, x); A             if(i!=N) { +Link (now+jia,now+m+i-1,1,0); theLink (now+jia,now+m+i,1,0); -             } $             ElseLink (now+jia,t1,1,0); the         } the      while(SPFA ()); theprintf"%d", ans); the     return 0; -}

The game problem of codevs1033 earthworms