（转载前请标明出处，谢谢）
打算来做一波TJOI2015，来写题解啦！
Day1:
T1:［bzoj3996］
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3996
首先我们对题目中的式子化简一下，得到
；
于是这就成了一个最小割模型：
- S和(i,j)连边，权值为b[i,j]
- (i,j)和i，j分别连边，权值为inf
- i和T连边，权值为c[i]
于是跑一下最小割就可以了；
（然而不知道发生了什么。。。最小割跑不过去。。。似乎bzoj把p党卡常数了QAQ）
（cyand神犇说他在省选的时候贪心了一发，就过了，至今找不出反例，于是我水过去了）
最小割代码如下：
1 uses math; 2 var s,t,n,i,j,tot,w,cnt:longint; 3 last,pre,other,flow:array[0..2600000] of longint; 4 l,q:array[0..3000000] of longint; 5 ans:int64; 6 procedure insert(a,b,c,d:longint); 7 begin 8 pre[tot]:=last[a]; 9 last[a]:=tot; 10 other[tot]:=b; 11 flow[tot]:=c; 12 inc(tot); 13 pre[tot]:=last[b]; 14 last[b]:=tot; 15 other[tot]:=a; 16 flow[tot]:=d; 17 inc(tot); 18 end; 19 function bfs:boolean; 20 var s1,t1,u,v,q1:longint; 21 begin 22 fillchar(q,sizeof(q),0); 23 for i:=0 to n do 24 l[i]:=-1; 25 s1:=0; 26 t1:=1; 27 l[s]:=1; 28 q[t1]:=s; 29 while (s1<t1) do 30 begin 31 inc(s1); 32 u:=q[s1]; 33 q1:=last[u]; 34 while (q1>=0) do 35 begin 36 v:=other[q1]; 37 if (flow[q1]>0) and (l[v]=-1) then 38 begin 39 inc(t1); 40 q[t1]:=v; 41 l[v]:=l[u]+1; 42 end; 43 q1:=pre[q1]; 44 end; 45 end; 46 if (l[t]=-1) then exit(false) else exit(true); 47 end; 48 function find(u,int:longint):longint; 49 var w,v,q1,t1:longint; 50 begin 51 if (u=t) then exit(int); 52 w:=0; 53 q1:=last[u]; 54 while (q1>=0) and (w<int) do 55 begin 56 v:=other[q1]; 57 if (l[v]=l[u]+1) then 58 begin 59 t1:=find(v,min(flow[q1],int-w)); 60 flow[q1]:=flow[q1]-t1; 61 flow[q1 xor 1]:=flow[q1 xor 1]+t1; 62 w:=w+t1; 63 end; 64 q1:=pre[q1]; 65 end; 66 if (w>=int) then l[u]:=-1; 67 exit(w); 68 end; 69 function dinic:int64; 70 var e:longint; 71 ans:int64; 72 begin 73 ans:=0; 74 while bfs do 75 begin 76 e:=find(s,maxlongint); 77 ans:=ans+e; 78 end; 79 exit(ans); 80 end; 81 begin 82 readln(n); 83 s:=n*n+n+1; 84 t:=n*n+n+2; 85 tot:=0; 86 for i:=0 to t do 87 last[i]:=-1; 88 cnt:=n; 89 ans:=0; 90 for i:=1 to n do 91 begin 92 for j:=1 to n do 93 begin 94 read(w); 95 inc(cnt); 96 insert(s,cnt,w,0); 97 insert(cnt,i,maxlongint,0); 98 if (i<>j) then insert(cnt,j,maxlongint,0); 99 ans:=ans+w; 100 end; 101 readln; 102 end; 103 for i:=1 to n do 104 begin 105 read(w); 106 insert(i,t,w,0); 107 end; 108 readln; 109 n:=t; 110 writeln(ans-dinic); 111 end. 112
贪心代码如下：
1 var n,i,j,s,ans,max,max1:longint; 2 b:array[0..1000,0..1000] of longint; 3 c,inc:array[0..1000] of longint; 4 begin 5 readln(n); 6 for i:=1 to n do 7 begin 8 for j:=1 to n do 9 read(b[i,j]); 10 readln; 11 end; 12 for i:=1 to n do 13 read(c[i]); 14 readln; 15 fillchar(inc,sizeof(inc),0); 16 s:=0; 17 ans:=0; 18 for i:=1 to n do 19 begin 20 s:=0; 21 for j:=1 to n do 22 s:=s+b[i,j]+b[j,i]; 23 inc[i]:=b[i,i]-s+c[i]; 24 end; 25 for i:=1 to n do 26 for j:=1 to n do 27 ans:=ans+b[i,j]; 28 for i:=1 to n do 29 ans:=ans-c[i]; 30 while true do 31 begin 32 max:=-1; 33 max1:=-1; 34 for i:=1 to n do 35 if (max<inc[i]) then 36 begin 37 max:=inc[i]; 38 max1:=i; 39 end; 40 if (max1=-1) then break; 41 ans:=ans+max; 42 for i:=1 to n do 43 inc[i]:=inc[i]+b[i,max1]+b[max1,i]; 44 inc[max1]:=-1; 45 end; 46 writeln(ans); 47 end. 48
T2:［bzoj3997］
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3997
本题要用一个Dilworth定理：DAG最小链覆盖＝最大独立子集，
于是发现最大独立子集显然符合题目，于是直接跑DP就可以了，方程如下：
f[i,j]=max{f[i-1,j+1]+a[i,j],f[i-1,j],f[i,j+1]}
代码如下：
1 var t,l,n,m,i,j:longint; 2 a,f:array[0..1005,0..1005] of int64; 3 begin 4 readln(t); 5 for l:=1 to t do 6 begin 7 readln(n,m); 8 fillchar(a,sizeof(a),0); 9 fillchar(f,sizeof(f),0); 10 for i:=1 to n do 11 begin 12 for j:=1 to m do 13 read(a[i,j]); 14 readln; 15 end; 16 for i:=1 to n do 17 for j:=m downto 1 do 18 begin 19 f[i,j]:=f[i-1,j+1]+a[i,j]; 20 if (f[i-1,j]>f[i,j]) then f[i,j]:=f[i-1,j]; 21 if (f[i,j+1]>f[i,j]) then f[i,j]:=f[i,j+1]; 22 end; 23 writeln(f[n,1]); 24 end; 25 end.
T3:［bzoj3998］
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3998
这题把我快写哭了QAQQQ。。。
这题是一个裸的后缀自动机，（然而我并不会，于是去看hzwer的博客）
于是写啊写啊写。。。然后狂WA不止。。。
问题在于，pascal党这里自动机建完以后千万不能写递归DFS。。。（C＋＋党随意）
pascal党可以改成非递归形式或者拓扑排序，就可以了QAQQQ。。。
终于AC了，撒花撒花！！！
代码如下：
1 var ch:array[0..500005] of char; 2 x:char; 3 n,i,j,tt,k,last,cnt,tot:longint; 4 a:array[0..1000005,0..25] of longint; 5 fa,mx,val,v,q,sum:array[0..1000005] of int64; 6 procedure extend(c:longint); 7 var p,np,q,nq,i,j:longint; 8 begin 9 p:=last; 10 inc(cnt); 11 last:=cnt; 12 np:=last; 13 mx[np]:=mx[p]+1; 14 val[np]:=1; 15 while (a[p,c]=0) and (p<>0) do 16 begin 17 a[p,c]:=np; 18 p:=fa[p]; 19 end; 20 if (p=0) then fa[np]:=1 21 else 22 begin 23 q:=a[p,c]; 24 if (mx[p]+1=mx[q]) then fa[np]:=q 25 else 26 begin 27 inc(cnt); 28 nq:=cnt; 29 mx[nq]:=mx[p]+1; 30 a[nq]:=a[q]; 31 fa[nq]:=fa[q]; 32 fa[q]:=nq; 33 fa[np]:=fa[q]; 34 while (a[p,c]=q) do 35 begin 36 a[p,c]:=nq; 37 p:=fa[p]; 38 end; 39 end; 40 end; 41 end; 42 procedure pre; 43 var i,j:longint; 44 t:int64; 45 begin 46 for i:=1 to cnt do 47 inc(v[mx[i]]); 48 for i:=1 to n do 49 v[i]:=v[i]+v[i-1]; 50 for i:=cnt downto 1 do 51 begin 52 q[v[mx[i]]]:=i; 53 dec(v[mx[i]]); 54 end; 55 for i:=cnt downto 1 do 56 begin 57 t:=q[i]; 58 if (tt=1) then val[fa[t]]:=val[fa[t]]+val[t] else val[t]:=1; 59 end; 60 val[1]:=0; 61 for i:=cnt downto 1 do 62 begin 63 t:=q[i]; 64 sum[t]:=val[t]; 65 for j:=0 to 25 do 66 sum[t]:=sum[t]+sum[a[t,j]]; 67 end; 68 end; 69 procedure dfs(x:longint); 70 var i:longint; 71 flag:boolean; 72 begin 73 while (k>val[x]) do 74 begin 75 k:=k-val[x]; 76 flag:=false; 77 for i:=0 to 25 do 78 begin 79 if (a[x,i]>0) and not(flag) then 80 begin 81 if (k<=sum[a[x,i]]) then 82 begin 83 write(chr(i+97)); 84 x:=a[x,i]; 85 flag:=true; 86 end 87 else 88 k:=k-sum[a[x,i]]; 89 end; 90 end; 91 end; 92 end; 93 begin 94 n:=0; 95 read(x); 96 while not((ord(x)>=48) and (ord(x)<=57)) do 97 begin 98 if (ord(x)>=97) and (ord(x)<=97+25) then 99 begin 100 inc(n); 101 ch[n]:=x; 102 end; 103 read(x); 104 end; 105 while not((ord(x)>=48) and (ord(x)<=57)) do read(x); 106 tt:=ord(x)-48; 107 read(x); 108 readln(k); 109 last:=1; 110 cnt:=1; 111 fillchar(fa,sizeof(fa),0); 112 fillchar(mx,sizeof(mx),0); 113 fillchar(val,sizeof(val),0); 114 fillchar(sum,sizeof(sum),0); 115 fillchar(v,sizeof(v),0); 116 fillchar(q,sizeof(q),0); 117 for i:=1 to n do 118 extend(ord(ch[i])-97); 119 pre; 120 tot:=0; 121 if (k>sum[1]) then write('-1') else dfs(1); 122 writeln; 123 end.
Day2:
(T1留个坑，之后补）
T2:[bzoj4000]
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=4000
这一题真心有毒QAQQQ
首先题目意思理解继续要读10遍题目QAQQQ注意行数是从0开始的QAQQQ
然后就是一个状态压缩DP。。。
显然过不了TATTT。。。
那就来一发矩阵乘法！
这就是正解了。。。然后作为pascal党的我木有过。。。这题bzoj上面pascal要用double（卡精度）。。。于是TLE。。。
不过介于cojs上面过了，所以一样贴过来了。
代码如下：
1 type arr=array[0..70,0..70] of int64; 2 var n,m,p,k,i,j,x,max:longint; 3 a:arr; 4 f:array[0..70] of longint; 5 mp:array[0..2,0..100] of longint; 6 ans,modp:int64; 7 function time(a:arr;b:longint):arr; 8 var ans,now,anss:arr; 9 i,j,k,r:longint; 10 begin 11 fillchar(ans,sizeof(ans),0); 12 for i:=0 to max do 13 ans[i,i]:=1; 14 now:=a; 15 r:=b; 16 while (r>0) do 17 begin 18 if (r mod 2=1) then 19 begin 20 fillchar(anss,sizeof(anss),0); 21 for i:=0 to max do 22 for j:=0 to max do 23 for k:=0 to max do 24 anss[i,j]:=(anss[i,j]+(int64(ans[i,k]*now[k,j]))); 25 ans:=anss; 26 end; 27 r:=r div 2; 28 fillchar(anss,sizeof(anss),0); 29 for i:=0 to max do 30 for j:=0 to max do 31 for k:=0 to max do 32 anss[i,j]:=anss[i,j]+(int64(now[i,k]*now[k,j])); 33 now:=anss; 34 end; 35 exit(ans); 36 end; 37 function tryit(a,b:longint):boolean; 38 var i,j,t,k:longint; 39 begin 40 for i:=0 to m-1 do 41 begin 42 if ((a shr i) and 1<>0) then 43 begin 44 for j:=1 to mp[1,0] do 45 begin 46 k:=i+mp[1,j]; 47 if (k>=0) and (k<=m-1) and ((a shl k) and 1<>0) then exit(false); 48 end; 49 for j:=1 to mp[2,0] do 50 begin 51 k:=i+mp[2,j]; 52 if (k>=0) and (k<=m-1) and ((b shr k) and 1<>0) then exit(false); 53 end; 54 end; 55 end; 56 for i:=0 to m-1 do 57 begin 58 if ((b shr i) and 1<>0) then 59 begin 60 for j:=1 to mp[1,0] do 61 begin 62 k:=i+mp[1,j]; 63 if (k>=0) and (k<=m-1) and ((b shr k) and 1<>0) then exit(false); 64 end; 65 for j:=1 to mp[0,0] do 66 begin 67 k:=i+mp[0,j]; 68 if (k>=0) and (k<=m-1) and ((a shr k) and 1<>0) then exit(false); 69 end; 70 end; 71 end; 72 exit(true); 73 end; 74 begin 75 modp:=1; 76 for i:=1 to 32 do 77 modp:=int64(modp*2); 78 readln(n,m); 79 readln(p,k); 80 max:=1 shl m; 81 dec(max); 82 fillchar(mp,sizeof(mp),0); 83 for i:=0 to 2 do 84 begin 85 for j:=0 to p-1 do 86 begin 87 read(x); 88 if (x=1) and not((i=1) and (j=k)) then 89 begin 90 inc(mp[i,0]); 91 mp[i,mp[i,0]]:=j-k; 92 end; 93 end; 94 end; 95 fillchar(a,sizeof(a),0); 96 fillchar(f,sizeof(f),0); 97 for i:=0 to max do 98 for j:=0 to max do 99 if tryit(i,j) then a[i,j]:=1 else a[i,j]:=0; 100 for i:=0 to max do 101 f[i]:=a[0,i]; 102 a:=time(a,n); 103 ans:=0; 104 for i:=0 to max do 105 begin 106 if (f[i]<>0) then ans:=(ans+a[i,0]) mod modp; 107 while (ans>=modp) do ans:=ans-modp; 108 while (ans<0) do ans:=ans+modp; 109 end; 110 writeln(ans); 111 end. 112
T3:[bzoj4001]
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=4001
这一题一定是这次省选最友善的一个题。。。
首先我们可以打暴力（或者手算），于是找到了规律，
可以用组合数学证明一发（然而我不会。。。）QAQQQ
总之就是个结论题QAQQQ
代码如下：
1 var n:longint; 2 x:real; 3 begin 4 readln(n); 5 x:=n/(2*n-1); 6 x:=x*(n+1)/2; 7 writeln(x:0:9); 8 end.