题意：给出一个矩阵前i列所有元素的和，和前j行所有元素的和，求这个矩阵解压以后的原型。（答案不唯一）
n,m<=20,1<=a[i,j]<=20
思路：这道题把边上的流量作为原先矩阵中的点
把每一行，每一列都看成一个点
S——>i行 a[i]-m
i行——>j列 19
j列——>T b[i]-n
跑最大流，边(i,j+n)上的流量就是a[i,j]的值
为什么容量是a[i]-m,19,b[i]-n？
因为点权（边权）不能为0，所以要先把所有值+1，上限就-1，输出的时候+1即可
1 var fan:array[1..200000]of longint; 2 head,vet,next,len,gap,dis,a,b,c1,c2:array[0..50000]of longint; 3 num:array[1..30,1..30]of longint; 4 n,m,i,j,tot,s,source,src,cas,v:longint; 5 6 procedure add(a,b,c:longint); 7 begin 8 inc(tot); 9 next[tot]:=head[a]; 10 vet[tot]:=b; 11 len[tot]:=c; 12 head[a]:=tot; 13 14 inc(tot); 15 next[tot]:=head[b]; 16 vet[tot]:=a; 17 len[tot]:=0; 18 head[b]:=tot; 19 end; 20 21 function min(x,y:longint):longint; 22 begin 23 if x<y then exit(x); 24 exit(y); 25 end; 26 27 function dfs(u,aug:longint):longint; 28 var e,v,t,flow,val:longint; 29 begin 30 if u=src then exit(aug); 31 e:=head[u]; flow:=0; val:=s-1; 32 while e<>0 do 33 begin 34 v:=vet[e]; 35 if len[e]>0 then 36 begin 37 if dis[u]=dis[v]+1 then 38 begin 39 t:=dfs(v,min(len[e],aug-flow)); 40 len[e]:=len[e]-t; 41 len[fan[e]]:=len[fan[e]]+t; 42 flow:=flow+t; 43 if dis[source]>=s then exit(flow); 44 if aug=flow then break; 45 end; 46 val:=min(val,dis[v]); 47 end; 48 e:=next[e]; 49 end; 50 if flow=0 then 51 begin 52 dec(gap[dis[u]]); 53 if gap[dis[u]]=0 then dis[source]:=s; 54 dis[u]:=val+1; 55 inc(gap[dis[u]]); 56 end; 57 exit(flow); 58 end; 59 60 function maxflow:longint; 61 var ans:longint; 62 begin 63 fillchar(gap,sizeof(gap),0); 64 fillchar(dis,sizeof(dis),0); 65 gap[0]:=s; ans:=0; 66 while dis[source]<s do ans:=ans+dfs(source,maxlongint); 67 exit(ans); 68 end; 69 70 begin 71 assign(input,'UVA11082.in'); reset(input); 72 assign(output,'UVA11082.out'); rewrite(output); 73 for i:=1 to 200000 do 74 if i mod 2=1 then fan[i]:=i+1 75 else fan[i]:=i-1; 76 readln(cas); 77 for v:=1 to cas do 78 begin 79 fillchar(head,sizeof(head),0); tot:=0; 80 fillchar(num,sizeof(num),0); 81 fillchar(len,sizeof(len),0); 82 readln(n,m); 83 for i:=1 to n do 84 begin 85 read(a[i]); c1[i]:=a[i]-a[i-1]; 86 end; 87 for i:=1 to m do 88 begin 89 read(b[i]); c2[i]:=b[i]-b[i-1]; 90 end; 91 s:=n+m+2; source:=n+m+1; src:=n+m+2; 92 for i:=1 to n do add(source,i,c1[i]-m); 93 for i:=1 to m do add(i+n,src,c2[i]-n); 94 for i:=1 to n do 95 for j:=1 to m do 96 begin 97 num[i,j]:=tot+2; 98 add(i,j+n,19); 99 end; 100 maxflow; 101 writeln('Matrix ',v); 102 for i:=1 to n do 103 begin 104 for j:=1 to m-1 do write(len[num[i,j]]+1,' '); 105 write(len[num[i,m]]+1); 106 writeln; 107 end; 108 if v<>cas then writeln; 109 end; 110 111 close(input); 112 close(output); 113 end.