好题，这道题可以用线段树来快速模拟费用流寻找最长增广路
这样修改怎么做也很显然了
1 type node=record 2 s,lx,rx,mx,lp,rp,pb,pe:longint; 3 end; 4 5 var tree:array[0..100010*4,0..1] of node; 6 rev:array[0..100010*4] of boolean; 7 a:array[0..100010] of longint; 8 q:array[0..100010] of node; 9 j,t,ans,i,n,m,x,y,k,ch:longint; 10 c:node; 11 12 procedure swap(var a,b:node); 13 var c:node; 14 begin 15 c:=a; 16 a:=b; 17 b:=c; 18 end; 19 20 procedure put(var a:node; x:longint); 21 begin 22 a.lx:=x; 23 a.rx:=x; 24 a.mx:=x; 25 a.s:=x; 26 end; 27 28 procedure update(var c:node; a,b:node); 29 begin 30 c.s:=a.s+b.s; 31 c.lx:=a.lx; c.lp:=a.lp; 32 if a.s+b.lx>c.lx then 33 begin 34 c.lx:=a.s+b.lx; 35 c.lp:=b.lp; 36 end; 37 c.rx:=b.rx; c.rp:=b.rp; 38 if b.s+a.rx>c.rx then 39 begin 40 c.rx:=b.s+a.rx; 41 c.rp:=a.rp; 42 end; 43 44 c.mx:=a.rx+b.lx; c.pb:=a.rp; c.pe:=b.lp; //这里要维护位置，相当于记录增广路的起点终点 45 if c.mx<a.mx then 46 begin 47 c.mx:=a.mx; 48 c.pb:=a.pb; 49 c.pe:=a.pe; 50 end; 51 if c.mx<b.mx then 52 begin 53 c.mx:=b.mx; 54 c.pb:=b.pb; 55 c.pe:=b.pe; 56 end; 57 end; 58 59 procedure build(i,l,r:longint); 60 var m,j:longint; 61 begin 62 if l=r then 63 begin 64 put(tree[i,1],a[l]); 65 put(tree[i,0],-a[l]); 66 for j:=0 to 1 do 67 begin 68 tree[i,j].lp:=l; 69 tree[i,j].rp:=l; 70 tree[i,j].pb:=l; 71 tree[i,j].pe:=l; 72 end; 73 end 74 else begin 75 m:=(l+r) shr 1; 76 build(i*2,l,m); 77 build(i*2+1,m+1,r); 78 update(tree[i,1],tree[i*2,1],tree[i*2+1,1]); 79 update(tree[i,0],tree[i*2,0],tree[i*2+1,0]); 80 end; 81 end; 82 83 procedure change(i:longint); 84 begin 85 swap(tree[i,0],tree[i,1]); 86 rev[i]:=not rev[i]; 87 end; 88 89 procedure push(i:longint); 90 begin 91 rev[i]:=false; 92 change(i*2); 93 change(i*2+1); 94 end; 95 96 procedure work(i,l,r:longint); 97 var m:longint; 98 begin 99 if l=r then 100 begin 101 put(tree[i,1],y); //一条边和它的反向弧 102 put(tree[i,0],-y); 103 end 104 else begin 105 m:=(l+r) shr 1; 106 if rev[i] then push(i); 107 if x<=m then work(i*2,l,m) 108 else work(i*2+1,m+1,r); 109 update(tree[i,1],tree[i*2,1],tree[i*2+1,1]); 110 update(tree[i,0],tree[i*2,0],tree[i*2+1,0]); 111 end; 112 end; 113 114 function ask(i,l,r:longint):node; 115 var m:longint; 116 s,s1,s2:node; 117 begin 118 if (x<=l) and (y>=r) then exit(tree[i,1]) 119 else begin 120 m:=(l+r) shr 1; 121 if rev[i] then push(i); 122 if y<=m then exit(ask(i*2,l,m)); 123 if x>m then exit(ask(i*2+1,m+1,r)); 124 s1:=ask(i*2,l,m); 125 s2:=ask(i*2+1,m+1,r); 126 update(s,s1,s2); 127 exit(s); 128 end; 129 end; 130 131 procedure rever(i,l,r,x,y:longint); //翻转，因为图的特殊性可知 132 var m:longint; 133 begin 134 if (x<=l) and (y>=r) then change(i) 135 else begin 136 m:=(l+r) shr 1; 137 if rev[i] then push(i); 138 if x<=m then rever(i*2,l,m,x,y); 139 if y>m then rever(i*2+1,m+1,r,x,y); 140 update(tree[i,1],tree[i*2,1],tree[i*2+1,1]); 141 update(tree[i,0],tree[i*2,0],tree[i*2+1,0]); 142 end; 143 end; 144 145 begin 146 readln(n); 147 for i:=1 to n do 148 read(a[i]); 149 build(1,1,n); 150 readln(m); 151 for i:=1 to m do 152 begin 153 read(ch); 154 if ch=1 then 155 begin 156 readln(x,y,k); 157 ans:=0; 158 t:=0; 159 for j:=1 to k do 160 begin 161 c:=ask(1,1,n); 162 if c.mx>0 then ans:=ans+c.mx 163 else break; 164 rever(1,1,n,c.pb,c.pe); 165 inc(t); 166 q[t]:=c; 167 end; 168 for j:=t downto 1 do 169 rever(1,1,n,q[j].pb,q[j].pe); 170 writeln(ans); 171 end 172 else begin 173 readln(x,y); 174 work(1,1,n); 175 end; 176 end; 177 end.
