wikioi1034 家园
这是一道神奇的网络流。具体详细的题解以及代码可以去Orz这个大神的博客。
传送门：http://blog.csdn.net/lytning/article/details/23034379
简单地说，就是要建立一个模型，就是把每个点（包括地球和月球）都分成无数多个点……比如第j个时刻的地球，就用地球i号表示，第i个中转站就用i中转站j表示。（处理是把月球-1自动改为n+1吧）。把飞船看成是边。然后就是神奇的跑一遍sap看是否能运那么多人，不行再扩展一层节点在跑一遍……orz语言能力不好还是去看上面那个大神的博客吧。最后把数组开大！把记录边的数组开到很大很大……不然就和我一样wa了好多好多次……或者把time控制小一点……比如等于100是就不要找了根本找不到的，这样时间差不多就剩下100ms（time为200要跑700ms）……
type arr=record toward,cap,next:longint; end;var edge:array[0..100000]of arr; d,num,first,cur,h:array[0..5000]of longint; c:array[0..50,0..10000]of longint; ss,tot,sum,st,s,t,i,j,k,n,m,kk,time,flow,leftk:longint; function cal(i,j:longint):longint; begin exit((n*2)*j+i+2); end; function min(x,y:longint):longint; begin if x>y then exit(y); exit(x); end; procedure add(i,j,k:longint); begin edge[tot].toward:=j; edge[tot].cap:=k; edge[tot].next:=first[i]; first[i]:=tot; inc(tot); end; procedure addedge(i,j,k:longint); begin add(i,j,k); add(j,i,0); end; function sap(v,flow:longint):longint; var rec,ret,i,j:longint; begin if v=st then exit(flow); rec:=0; i:=cur[v]; while i<>-1 do begin j:=edge[i].toward; if (edge[i].cap>0) and (d[v]=d[j]+1) then begin ret:=sap(j,min(flow-rec,edge[i].cap)); dec(edge[i].cap,ret); inc(edge[i xor 1].cap,ret); cur[v]:=i; inc(rec,ret); if rec=flow then exit(flow); end; i:=edge[i].next; end; dec(num[d[v]]); if num[d[v]]=0 then d[ss]:=sum; inc(d[v]); inc(num[d[v]]); cur[v]:=first[v]; exit(rec); end; function maxflow:longint; var flow:longint; begin fillchar(num,sizeof(num),0); fillchar(d,sizeof(d),0); for i:=0 to sum do cur[i]:=first[i]; num[0]:=sum; flow:=0; while d[ss]<sum do inc(flow,sap(ss,maxlongint)); exit(flow); end; procedure more(time:longint); var i,j,k:longint; begin inc(sum,n+2); inc(num[0],n+2); if time=0 then addedge(ss,cal(s,time),kk); addedge(cal(t,time),st,maxlongint); if time>0 then begin for i:=s to n do addedge(cal(i,time-1),cal(i,time),maxlongint); addedge(cal(t,time-1),cal(t,time),maxlongint); for i:=1 to m do begin j:=time mod c[i,0]+1; k:=time mod c[i,0]; if k=0 then k:=c[i,0]; addedge(cal(c[i,k],time-1),cal(c[i,j],time),h[i]); end; end; end; begin readln(n,m,kk); leftk:=kk; fillchar(edge,sizeof(edge),0); fillchar(num,sizeof(num),0); for i:=0 to 200*(n+2) do first[i]:=-1; tot:=0; s:=0; t:=n+1; ss:=0; st:=1; sum:=2; num[0]:=2; for i:=1 to m do begin read(h[i],c[i,0]); for j:=1 to c[i][0] do begin read(c[i,j]); if c[i,j]=-1 then c[i,j]:=n+1; end; readln; end; time:=0; while time<200 do begin more(time); flow:=0; inc(flow,maxflow); dec(leftk,flow); if leftk<=0 then begin writeln(time); break; end; inc(time); end; if time=200 then writeln('0'); end.
vijos 1653 疯狂方格取数
zkw处理负边上跪了很久。网上少有资料
const maxn= 100000; type edgetype=record toward,cap,cost,next:longint; end; var edge:array[0..maxn] of edgetype; map:array[1..1000,1..1000,0..1] of longint; q,first,slack,d,dist:array[0..maxn] of longint; pd:array[0..maxn] of boolean; i,j,n,m,ss,st,tot,k,x,cnt,point,mincost:longint; function min(x,y:longint):longint; begin if x<y then exit(x) else exit(y); end; procedure add(i,j,k,l:longint); begin edge[tot].toward:=j; edge[tot].cap:=k; edge[tot].cost:=l; edge[tot].next:=first[i]; first[i]:=tot; inc(tot); end; procedure addedge(i,j,k,l:longint); begin add(i,j,k,l); add(j,i,0,-l); end; procedure spfa; var head,tail,i,j,tmp,value:longint; begin fillchar(pd,sizeof(pd),false); for i:= ss to st do d[i]:=maxlongint; head:=1; tail:=1; q[1]:=ss; pd[ss]:=true; d[ss]:=0; while head<=tail do begin j:=q[head]; i:=first[j]; while i<>-1 do begin tmp:=edge[i].toward; value:=edge[i].cost; if (edge[i].cap>0) and (d[tmp]>d[j]+value) then begin d[tmp]:=d[j]+value; if not pd[tmp] then begin inc(tail); if tail=maxn then tail:=1; q[tail]:=tmp; pd[tmp]:=true; end; end; i:=edge[i].next; end; inc(head); if head=maxn then head:=1; pd[j]:=false; end; end; procedure dfs(v:longint); var i,value,tmp:longint; begin pd[v]:=true; i:=first[v]; while i<>-1 do begin tmp:=edge[i].toward; value:=edge[i].cost; if (edge[i].cap>0) and (not pd[tmp]) and (d[tmp]=d[v]+value) then begin dist[tmp]:=dist[v]-value; dfs(tmp); end; i:=edge[i].next; end; end; function aug(v,flow:longint):longint; var rec,ret,i,tmp,value:longint; begin if v=st then begin inc(mincost,flow*(dist[st]-dist[ss])); exit(flow); end; rec:=0; i:=first[v]; pd[v]:=true; while i<>-1 do begin tmp:=edge[i].toward; value:=edge[i].cost; if (not pd[tmp]) and (edge[i].cap>0) then if dist[v]=dist[tmp]+value then begin ret:=aug(tmp,min(flow-rec,edge[i].cap)); dec(edge[i].cap,ret); inc(edge[i xor 1].cap,ret); inc(rec,ret); if rec=flow then exit(flow); end else slack[tmp]:=min(slack[tmp],dist[tmp]+value-dist[v]); i:=edge[i].next; end; exit(rec); end; function relabel:boolean; var i,delta:longint; begin delta:=maxlongint; for i:= ss to st do if not pd[i] then delta:=min(delta,slack[i]); if delta=maxlongint then exit(false); for i:= ss to st do if pd[i] then inc(dist[i],delta); exit(true); end; procedure init; begin readln(k,m,n); cnt:=m*n; ss:=0; st:=cnt*2+1; tot:=0; point:=0; for i:= ss to st do first[i]:=-1; for i:=1 to n do for j:= 1 to m do begin read(x); inc(point);map[i,j,0]:=point; inc(point);map[i,j,1]:=point; addedge(map[i,j,0],map[i,j,1],1,-x); addedge(map[i,j,0],map[i,j,1],maxlongint,0); end; addedge(ss,map[1,1,0],k,0); addedge(map[n,m,1],st,k,0); for i:=1 to n do for j:= 1 to m do begin if i+1<=n then addedge(map[i,j,1],map[i+1,j,0],maxlongint,0); if j+1<=m then addedge(map[i,j,1],map[i,j+1,0],maxlongint,0); end; end; procedure costflow; begin spfa; fillchar(pd,sizeof(pd),false); fillchar(dist,sizeof(dist),0); dfs(ss); repeat for i:= ss to st do slack[i]:=maxlongint; repeat fillchar(pd,sizeof(pd),false); until aug(ss,maxlongint)<=0; until not relabel; end; Begin init; costflow; writeln(mincost); end.
vijos 1726 美食节（费用流）
很明显是一道费用流，如果用常规的方法，把每个厨师每个时刻拆出来的话，由于点太多会爆于是我们考虑了这样一个事实，每个厨师并不需要把他每个时刻的点都扩展出来，一个厨师只有当他做完一个菜是才去考虑让他多做一个菜，这样的话由于一种只有sum(=∑p[i])道菜，所有整个图实际需要的点数就是源点汇点（2个），菜（n个），厨师（m个）以及扩展出来的厨师（sum个，实际只需sum-m，但为了方法我们不妨多几个点）。一开始源点连每道菜连一条容量为菜需求量费用为0的边，汇点和每个厨师连一条容量为1费用为0的边，然后把第i个厨师与第j道菜连一条容量为1费用为厨师做这道菜的时间的边。这样就是一个初始图。跑spfa（由于一次一定是做一道菜，所以就可以省点记录此次增广的流量），然后每次增广后必然会有一个厨师做出了某道菜，这是我们就扫一遍判断是哪个厨师做了菜（开个last数组记录每个厨师当前点与汇点的边，判断哪个厨师的最后一条边容量为0），然后开个chef表示这个厨师至今做了多少道菜。重新增加一个点，这个点与每道菜连一条容量为1费用为（这个厨师做这道菜的时间*这个厨师一共做了多少道菜），然后这个点再与汇点连一条容量为1费用为0的边。开个time计菜，等time=sum时就说明做完了，输出这时的费用就是答案了。
const maxn=300000; type edgetype=record toward,cap,cost,next:longint; end; var edge:array[0..maxn] of edgetype; minflow,pre,first,q,dist:array[0..maxn] of longint; pd:array[0..maxn] of boolean; p:array[1..40] of longint; data:array[1..40,1..100] of longint; chef:array[1..100] of longint; last:array[1..100] of longint; i,j,s,t,n,m,tot,cnt,time,x:longint; function min(x,y:longint):longint; begin if x<y then exit(x) else exit(y); end; procedure add(i,j,k,l:longint); begin edge[tot].toward:=j; edge[tot].cap:=k; edge[tot].cost:=l; edge[tot].next:=first[i]; first[i]:=tot; inc(tot); end; procedure addedge(i,j,k,l:longint); begin add(i,j,k,l); add(j,i,0,-l); end; function spfa:boolean; var tmp,value,i,j,head,tail:longint; begin fillchar(pd,sizeof(pd),false); for i:=0 to t do dist[i]:=maxlongint; head:=1; tail:=1; q[tail]:=s; pd[s]:=true; dist[s]:=0; while head<=tail do begin j:=q[head]; i:=first[j]; while i<>-1 do begin tmp:=edge[i].toward; value:=edge[i].cost; if (edge[i].cap>0) and (dist[tmp]>dist[j]+value) then begin dist[tmp]:=dist[j]+value; pre[tmp]:=i; if not pd[tmp] then begin inc(tail); q[tail]:=tmp; pd[tmp]:=true; end; end; i:=edge[i].next; end; inc(head); pd[j]:=false; end; if dist[t]=maxlongint then exit(false) else exit(true); end; function mcmf:longint; var u,cost:longint; begin cost:=0; while time<cnt do begin if not spfa then break; u:=t; while u<>s do begin dec(edge[pre[u]].cap); inc(edge[pre[u] xor 1].cap); u:=edge[pre[u] xor 1].toward; end; inc(cost,dist[t]); inc(time); for i:= 1 to m do if edge[last[i]].cap=0 then begin x:=i; break; end; inc(chef[x]); for i:=1 to n do addedge(i,n+m+time,1,data[i,x]*chef[x]); addedge(n+m+time,t,1,0); last[x]:=tot-2; end; exit(cost); end; procedure init; begin tot:=0; time:=0; cnt:=0; read(n,m); for i:=1 to n do begin read(p[i]); inc(cnt,p[i]); end; for i:= 1 to n do for j:= 1 to m do read(data[i,j]); s:=n+m+cnt+1; t:=s+1; for i:= 0 to t do first[i]:=-1; for i:=1 to n do addedge(s,i,p[i],0); for i:=1 to n do for j:= 1 to m do addedge(i,j+n,1,data[i,j]); for i:= 1 to m do chef[i]:=1; for i:=1 to m do begin addedge(n+i,t,1,0); last[i]:=tot-2; end; end; Begin init; writeln(mcmf); End.
vijos 1621 终极情报网（费用流）
网上没有太多题解，因为这本来就是一道很简单的模型，这么简单就不想说了，这道题主要难在两个问题，一个是要保留五位有效数字，一个是关于浮点运算（后面这个东西害得我两节晚自修没了，网上也没大神说一下，第一次遇到浮点运算什么的好讨厌）。前一种我很傻很天真的方法yy然后就完了，但是很难看，下面会给出某蔡大神的空间自行过去研究他的吧。关于第二个问题，这题是求一些实数的运算，可以用两个方法，一个是直接乘，就是把原来的spfa中+号改为*号，那么相应的，建边的时候反向弧的费用就不是-asi而是1/asi，这样就会遇到一个精度问题，所有在增广时必须要把原来的dist[to]<dist[i]*edge[j].cost改为dist[i]*edge[j].cost-dist[to]＞esp（esp=0.000000001）。这样程序才跑得动……第二种方法就是取对数，把*改为+，觉得神奇到不可思议，所以还是由蔡大神的题解来告诉我们这些蒟蒻吧。
type arr=record toward,next,cap:longint; cost:double; end; var edge:array[0..1000000]of arr; dist,minc,a:array[0..100000]of double; q,first,pre,minf,b:array[0..100000]of longint; map:array[0..2000,0..2000]of double; i,j,k,l,s,t,tot,kk,maxflow,n,x:longint; chose:array[0..10000]of boolean; spent,h:double; function exp(x:double;y:longint):double; //快速幂递归版！yy一下就知道为什么用快速幂了 begin if y=1 then exit(x); exit(exp(x,y div 2)*exp(x,y-(y div 2))); end; function min(x,y:longint):longint; begin if x>y then exit(y); exit(x); end; procedure add(i,j,k:longint;l:double); begin edge[tot].toward:=j; edge[tot].cap:=k; edge[tot].cost:=l; edge[tot].next:=first[i]; first[i]:=tot; inc(tot); end; procedure addedge(i,j,k:longint;l:double); begin add(i,j,k,l); add(j,i,0,1/l); end; function spfa:boolean; var i,head,tail,flow,j,too:longint; value,mm:double; begin fillchar(chose,sizeof(chose),false); fillchar(dist,sizeof(dist),0); dist[s]:=1; head:=1; tail:=1; q[1]:=s; mm:=0.000000001; chose[s]:=true; minf[s]:=maxlongint; while head<=tail do begin i:=q[head]; j:=first[i]; while j<>-1 do begin too:=edge[j].toward; value:=edge[j].cost; if (edge[j].cap>0) and (dist[i]*value-dist[too]>mm) then begin dist[too]:=dist[i]*value; pre[too]:=j; minf[too]:=min(minf[i],edge[j].cap); minc[too]:=edge[j].cost; if not chose[too] then begin chose[too]:=true; inc(tail); q[tail]:=too; end; end; j:=edge[j].next; end; inc(head); chose[i]:=false; end; if dist[t]=0 then exit(false); exit(true); end; procedure mcmf; var j,k,v:longint; begin spent:=1; maxflow:=0; while spfa do begin j:=minf[t]; inc(maxflow,j); v:=t; while v<>s do begin spent:=spent*exp(minc[v],j); k:=pre[v]; dec(edge[k].cap,j); inc(edge[k xor 1].cap,j); v:=edge[k xor 1].toward; end; end; end; procedure outo; var ans,i,j,k:longint; begin {writeln(spent);} spent:=spent*10; i:=1; x:=trunc(spent); while x=0 do begin spent:=spent*10; inc(i); x:=trunc(spent); end; spent:=spent/10; j:=0; while j<=5 do begin inc(j); spent:=spent*10; b[j]:=trunc(spent); spent:=spent-b[j]; end; if b[6]>=5 then inc(b[5]); k:=5; while (b[k]>=10) and (k>1) do begin inc(b[k-1]); b[k]:=b[k]-10; end; if b[1]>9 then dec(i); write('0.'); for k:=1 to i-1do write(0); for k:=1 to 5 do write(b[k]); readln; readln; readln; end; begin readln(n,kk); fillchar(map,sizeof(map),0); t:=n+2; s:=n+1; tot:=0; for i:=0 to n+3 do first[i]:=-1; for i:=1 to n do read(a[i]); for i:=1 to n do begin read(b[i]); if b[i]>0 then addedge(s,i,b[i],a[i]); end; readln; for i:=1 to n do begin read(x); if x>0 then addedge(i,t,maxlongint,1); end; readln; read(j,k); while (j<>-1) and (k<>-1) do begin readln(h,l); addedge(j,k,l,h); addedge(k,j,l,h); read(j,k); end; addedge(t,t+1,kk,1); inc(t); mcmf; if maxflow<kk then writeln('0') else outo; end.
vijos 1525 生命之泉 （费用流）
这道题和志愿者招募那道题差不多，属于一个类型的，志愿者招募那道题网上可以找个很多好的题解！关于这类题简单的说就是写一些方程，然后会发现一些规律
最核心的就是：
一开始）P[1]=X[1]≥2
P[2]=X[1]+X[2]≥3
P[3]=X[2]+X[3]≥2
可变为）
P[1]=X[1]-Y[1]=2
P[2]=X[1]+X[2]-Y[2]=3
P[3]=X[2]+X[3]-Y[3]=2 （Y就是一个变量就对了）
然后我们在最前面加一个 P[0]=0的式子，用下一个式子减上一个式子就可以得到
P[1]-P[0]=X[1]-Y[1]=2
P[2]-P[1]=X[2]+Y[1]-Y[2]=1
P[3]-P[2]=-X[1]+X[3]+Y[2]-Y[3]=-1
P[4]-P[3]=-X[2]-X[3]+Y[3]=-2
将常数项左移，得
P[1]-P[0]=X[1]-Y[1]-2=0
P[2]-P[1]=X[2]+Y[1]-Y[2]-1=0
P[3]-P[2]=-X[1]+X[3]+Y[2]-Y[3]+1=0
P[4]-P[3]=-X[2]-X[3]+Y[3]+2=0
我们发现一个x或y只会在两个式子里面出现，而且一个是负的，一个是正的！（大神话：很容易联想到网络流）
对于每个x[i]，从-x[i]想+x[i]连一条边（具体要根据题意），对于每个y[i]也是一样，不过y[i]和-y[i]总是出现在一前一后。
这道题比较麻烦……要把最大费用最大流然后转为最小费用最大流跑zkw……
type arr=record toward,cap,cost,next:longint; end; const maxn=1000000; var edge:array[0..10000]of arr; first,d,dist,q,slack:array[0..100000]of longint; chose:array[0..100000]of boolean; i,j,k,l,s,t,n,m,tot,maxcost,maxflow,kk:longint; procedure add(i,j,k,l:longint); begin edge[tot].toward:=j; edge[tot].cap:=k; edge[tot].cost:=l; edge[tot].next:=first[i]; first[i]:=tot; inc(tot); end; procedure addedge(i,j,k,l:longint); begin add(i,j,k,l); add(j,i,0,-l); end; function min(x,y:longint):longint; begin if x>y then exit(y); exit(x); end; procedure spfa; var head,tail,i,too,value:longint; begin fillchar(chose,sizeof(chose),false); for i:=0 to t do d[i]:=maxlongint; head:=1; tail:=1; q[1]:=s; chose[s]:=true; d[s]:=0; while head<=tail do begin j:=q[head]; i:=first[j]; while i<>-1 do begin too:=edge[i].toward; value:=edge[i].cost; if (edge[i].cap>0) and (d[too]>d[j]+value) then begin d[too]:=d[j]+value; if not chose[too] then begin inc(tail); if tail=maxn then tail:=1; q[tail]:=too; chose[too]:=true; end; end; i:=edge[i].next; end; inc(head); if head=maxn then head:=1; chose[j]:=false; end; end; procedure dfs(v:longint); var i,value,too:longint; begin chose[v]:=true; i:=first[v]; while i<>-1 do begin too:=edge[i].toward; value:=edge[i].cost; if (edge[i].cap>0) and (not chose[too]) and (d[too]=d[v]+value) then begin dist[too]:=dist[v]-value; dfs(too); end; i:=edge[i].next; end; end; function aug(v,flow:longint):longint; var rec,ret,too,value,i:longint; begin if v=t then begin inc(maxcost,(dist[t]-dist[s])*flow); inc(maxflow,flow); exit(flow); end; i:=first[v]; chose[v]:=true; rec:=0; while i<>-1 do begin too:=edge[i].toward; value:=edge[i].cost; if (edge[i].cap>0) and (not chose[too]) then if dist[v]=dist[too]+value then begin ret:=aug(too,min(flow-rec,edge[i].cap)); dec(edge[i].cap,ret); inc(edge[i xor 1].cap,ret); inc(rec,ret); if rec=flow then exit(flow); end else slack[too]:=min(slack[too],dist[too]+value-dist[v]); i:=edge[i].next; end; exit(rec); end; function rel:boolean; var spent,i:longint; begin spent:=maxlongint; for i:=0 to t do if not chose[i] then spent:=min(spent,slack[i]); if spent=maxlongint then exit(false); for i:=0 to t do if chose[i] then inc(dist[i],spent); exit(true); end; procedure costflow; begin spfa; fillchar(chose,sizeof(chose),false); fillchar(dist,sizeof(dist),0); dfs(s); repeat for i:=0 to t do slack[i]:=maxlongint; repeat fillchar(chose,sizeof(chose),false); until aug(s,maxlongint)<=0; until not rel; end; procedure into; begin readln(n,m,kk); s:=0; t:=n+2; tot:=0; for i:=0 to t do first[i]:=-1; addedge(1,t,kk,0); addedge(s,n+1,kk,0); for i:=1 to m do begin readln(j,k,l); addedge(k+1,j,1,-l); end; for i:=1 to n do addedge(i+1,i,maxlongint,0); fillchar(d,sizeof(d),0); end; begin into; costflow; writeln(maxcost); readln; readln; end.
vijos 1499 炸毁燃料库
建边一直跪……为了容易想容易看，结果竟然就跪了……
后来认真想了好久（Orz蔡大神的代码）才终于过了……线性规划
一开始
x[1]+x[2]+x[3]+...+x[m] <=k;
x[2]+x[3]+x[4]+...+x[m+1] <=k;
x[3]+x[4]+x[5]+...+x[m+2] <=k;
......
x[n-m+1]+...+x[n] <=k;
转换为
0=0
x[1]+x[2]+x[3]+...+x[m] -y[1]-k=0;
x[2]+x[3]+x[4]+...+x[m+1] -y[2]-k=0;
x[3]+x[4]+x[5]+...+x[m+2] -y[3]-k=0;
...
x[n-m+1]+...+x[n] -y[n-m+1]-k=0;
下面减上面
1 x[1]+x[2]+x[3]+...+x[m] -y[1]-k=0
2 x[m+1] -x[1] +y[1]-y[2] =0;
3 x[m+2] -x[2] +y[2]-y[3] =0;
4 x[m+3] -x[3] +y[3]-y[4] =0;
......
（ x[m+m] -x[m] +y[m]-y[m+1]=0
x[m+m+1]-x[m+1] ）
......
n-m+1 x[n] -x[n-m] +y[n-m]-y[n-m+1]=0;
n-m+2 -x[n-m+1]-x[n-m]...-x[n]+y[n-m+1]+k=0
然后就是中间那种情况会导致图有两种结果，判断条件是m≥n-m 是的话就是第一种图（没有括号那种情况），不是的话就是（有括号的图）
type arr=record toward,next,cap,cost:longint; end; const mm=1000000; var edge:array[0..100000]of arr; first,dist,d,q,slack,a:array[0..mm]of longint; i,j,k,l,n,m,s,t,tot,maxcost,maxflow,kk:longint; chose:array[0..mm]of boolean; procedure add(i,j,k,l:longint); begin edge[tot].toward:=j; edge[tot].cap:=k; edge[tot].cost:=l; edge[tot].next:=first[i]; first[i]:=tot; inc(tot); end; procedure addedge(i,j,k,l:longint); begin add(i,j,k,l); add(j,i,0,-l); end; function min(x,y:longint):longint; begin if x>y then exit(y); exit(x); end; procedure spfa; var head,tail,i,too,value:longint; begin fillchar(chose,sizeof(chose),false); for i:=0 to t do d[i]:=maxlongint; head:=1; tail:=1; q[1]:=s; chose[s]:=true; d[s]:=0; while head<=tail do begin j:=q[head]; i:=first[j]; while i<>-1 do begin too:=edge[i].toward; value:=edge[i].cost; if (edge[i].cap>0) and (d[too]>d[j]+value) then begin d[too]:=d[j]+value; if not chose[too] then begin inc(tail); if tail=mm then tail:=1; q[tail]:=too; chose[too]:=true; end; end; i:=edge[i].next; end; inc(head); if head=mm then head:=1; chose[j]:=false; end; end; procedure dfs(v:longint); var i,value,too:longint; begin chose[v]:=true; i:=first[v]; while i<>-1 do begin too:=edge[i].toward; value:=edge[i].cost; if (edge[i].cap>0) and (d[too]=d[v]+value) then begin dist[too]:=dist[v]-value; dfs(too); end; i:=edge[i].next; end; end; function aug(v,flow:longint):longint; var rec,ret,i,too,value:longint; begin if v=t then begin inc(maxcost,(dist[t]-dist[s])*flow); inc(maxflow,flow); exit(flow); end; rec:=0; chose[v]:=true; i:=first[v]; while i<>-1 do begin too:=edge[i].toward; value:=edge[i].cost; if (edge[i].cap>0) and (not chose[too]) then if dist[v]=dist[too]+value then begin ret:=aug(too,min(flow-rec,edge[i].cap)); dec(edge[i].cap,ret); inc(edge[i xor 1].cap,ret); inc(rec,ret); if rec=flow then exit(flow); end else slack[too]:=min(slack[too],dist[too]+value-dist[v]); i:=edge[i].next; end; exit(rec); end; function rel:boolean; var spent,i:longint; begin spent:=maxlongint; for i:=0 to t do if not chose[i] then spent:=min(spent,slack[i]); if spent=maxlongint then exit(false); for i:=0 to t do if chose[i] then inc(dist[i],spent); exit(true); end; procedure maxc; begin spfa; fillchar(chose,sizeof(chose),false); fillchar(dist,sizeof(dist),0); dfs(s); repeat for i:=0 to t do slack[i]:=maxlongint; repeat fillchar(chose,sizeof(chose),false); until aug(s,maxlongint)<=0; until not rel; end; procedure into; begin readln(n,m,kk); s:=0; t:=n-m+3; for i:=0 to t do first[i]:=-1; tot:=0; for i:=1 to n do read(a[i]); addedge(s,n-m+2,kk,0); addedge(1,t,kk,0); if m>=n-m then begin for i:=1 to n-m do addedge(i+1,1,1,-a[i]); for i:=n-m+1 to m do addedge(n-m+2,1,1,-a[i]); for i:=m+1 to n do addedge(n-m+2,i-m+1,1,-a[i]); end else begin for i:=1 to m do addedge(i+1,1,1,-a[i]); for i:=m+1 to n-m do addedge(i+1,i-m+1,1,-a[i]); for i:=n-m+1 to n do addedge(n-m+2,i-m+1,1,-a[i]); end; for i:=1 to n-m+1 do addedge(i+1,i,maxlongint,0); fillchar(d,sizeof(d),0); end; begin into; maxc; writeln(maxcost); end.
vijos 1734 海拔 （网络流转为最短路）
orz这题特别神奇，作为蒟蒻完全被考倒了。
找了网上的题解最好的就是这个了http://blog.sina.com.cn/s/blog_86942b1401014ajk.html （其实有另一个……但是莫名挂了，orz）。
这题很容易想到一个最小割模型，但是会tle！所以只能转为对偶图，这个神奇的东西推荐去看集训队论文：周冬《两极相通——浅析最大—最小定理在信息学竞赛中的应用》。讲的很详细！
然后就是spfa，裸的spfa不能过，所以要加lll+slt优化！
或者是二叉堆优化的dij……
然后这题由于信息量太大就跪了很久……先搞懂一系列概念，然后就是略蛋疼的spfa了，在tle无数次后终于把spfa改的能过最后一个点（一直卡最后一个点orz）……
//spfa版 type arr=record toward,next,cost:longint; end; const mm=300000; var edge:array[0..5*mm]of arr; first,d,q:array[0..mm]of longint; map:array[0..2000,0..2000]of longint; f:array[0..mm]of boolean; i,j,k,l,n,m,tot,s,t,sum:longint; procedure addedge(i,j,k:longint); begin inc(tot); edge[tot].toward:=j; edge[tot].next:=first[i]; first[i]:=tot; edge[tot].cost:=k; end; procedure spfa(v:longint); var head,tail,i,too,j,value,len,k:longint; sum:int64; begin sum:=0; len:=1; fillchar(f,sizeof(f),false); fillchar(d,sizeof(d),$7f); d[v]:=0; head:=0; tail:=1; q[1]:=v; f[v]:=true; while head<>tail do begin inc(head); if head>mm then head:=1; while d[q[head]]>sum div len do begin //lll优化，记得sum要int64！因为这个跪了一节晚自修！ inc(tail); if tail>mm then tail:=1; q[tail]:=q[head]; inc(head); if head>mm then head:=1; end; j:=q[head]; i:=first[j]; dec(len); dec(sum,d[j]); while i<>-1 do begin value:=edge[i].cost; too:=edge[i].toward; if (d[j]+value<d[too]) then begin d[too]:=d[j]+value; if (not f[too]) and (d[too]<d[t])then begin //d[too]<d[t] 这步很神奇！ inc(len); inc(sum,d[too]); k:=head+1; if k>mm then k:=1; if d[too]<d[q[k]] then begin //slf优化 f[too]:=true; q[head]:=too; head:=head-1; if head<1 then head:=mm; end else begin f[too]:=true; inc(tail); if tail>mm then tail:=1; q[tail]:=too; end; end; end; i:=edge[i].next; end; f[j]:=false; end; end; procedure into; begin s:=0; t:=n*n+1; sum:=0; tot:=0; for i:=0 to n do map[0,i]:=t; for i:=0 to n do map[i,n+1]:=t; for i:=0 to n do map[i,0]:=s; for i:=0 to n do map[n+1,i]:=s; for i:=1 to n do for j:=1 to n do begin inc(sum); map[i,j]:=sum; end; for i:=1 to n+1 do for j:=1 to n do begin read(k); addedge(map[i,j],map[i-1,j],k); end; for i:=1 to n do for j:=1 to n+1 do begin read(k); addedge(map[i,j-1],map[i,j],k); end; for i:=1 to n+1 do for j:=1 to n do begin read(k); addedge(map[i-1,j],map[i,j],k); end; for i:=1 to n do for j:=1 to n+1 do begin read(k); addedge(map[i,j],map[i,j-1],k); end; end; procedure outo; begin readln(n); for i:=0 to n*n+2 do first[i]:=-1; end; begin outo; into; spfa(s); writeln(d[t]); end. //据说这题dij快，然后就写了下，果然快了很多&…… type arr=record toward,next,cost:longint; end; ar=record value,toward:longint; end; const mm=300000; var edge:array[0..5*mm]of arr; first,g:array[0..mm]of longint; head:array[0..5*mm]of ar; map:array[0..2000,0..2000]of longint; f:array[0..mm]of boolean; i,j,k,l,n,m,tot,s,t,sum,len:longint; procedure addedge(i,j,k:longint); begin inc(tot); edge[tot].toward:=j; edge[tot].next:=g[i]; g[i]:=tot; edge[tot].cost:=k; end; procedure into; begin readln(n); for i:=0 to n*n+2 do g[i]:=-1; s:=n*n+1; t:=n*n+2; sum:=0; tot:=0; for i:=0 to n do map[0,i]:=t; for i:=0 to n do map[i,n+1]:=t; for i:=0 to n do map[i,0]:=s; for i:=0 to n do map[n+1,i]:=s; for i:=1 to n do for j:=1 to n do begin inc(sum); map[i,j]:=sum; end; for i:=1 to n+1 do for j:=1 to n do begin read(k); addedge(map[i,j],map[i-1,j],k); end; for i:=1 to n do for j:=1 to n+1 do begin read(k); addedge(map[i,j-1],map[i,j],k); end; for i:=1 to n+1 do for j:=1 to n do begin read(k); addedge(map[i-1,j],map[i,j],k); end; for i:=1 to n do for j:=1 to n+1 do begin read(k); addedge(map[i,j],map[i,j-1],k); end; end; procedure swap(a,b:longint); var j:ar; begin j:=head[a]; head[a]:=head[b]; head[b]:=j; first[head[a].toward]:=a; first[head[b].toward]:=b; end; procedure decre(i:longint); begin while (i<>1) and (head[i].value<head[i shr 1].value) do begin swap(i,i shr 1); i:=i shr 1; end; end; procedure headpify; var i:longint; begin i:=2; while i<=len do begin if (i<len) and (head[i+1].value<head[i].value) then inc(i); if head[i].value<head[i shr 1].value then begin swap(i,i shr 1); i:=i*2; end else break; end; end; procedure relax(i,j,k:longint); begin if k+head[first[i]].value<head[first[j]].value then begin head[first[j]].value:=k+head[first[i]].value; decre(first[j]); end; end; procedure dij; var i,too,va:longint; begin for i:=1 to t do begin head[i].toward:=i; head[i].value:=maxlongint; first[i]:=i; end; head[s].value:=0; swap(1,s); len:=t; while len>0 do begin i:=head[1].toward; if i=t then exit; swap(1,len); dec(len); headpify; j:=g[i]; while j<>-1 do begin too:=edge[j].toward; va:=edge[j].cost; if first[too]<=len then relax(i,too,va); j:=edge[j].next; end; end; end; begin into; dij; writeln(head[1].value); end.