摘要：典型动态规划，求最大子矩阵和，其实是最长子段和的变形，设f[i,j,k]表示从第i到j行第k列的和program poj1050;var a:array[1..100,1..100] of integer; f:array[0..100,0..100,0..100] of integer; n,i,j,x,k:integer; ans,s:longint;begin readln(n); x:=1; for i:=1 to n*n do begin read(k); if i mod n=0 then begin a[x,n]:=k; inc(x); end else a[x,i mod n] 阅读全文

摘要：求矩形中长为w，宽为h的最大子矩阵和，f[i,j]表示从（1，1）到（i，j）的和，f[i,j]=f[i,j]+f[i-1,j]+f[i,j-1]-f[i-1,j-1]ans=max(ans,f[i,j]-f[i-w,j]-f[i,j-h]+f[i-w,j-h]program poj2029;var f:array[0..100,0..100] of integer; n,i,j,x,y,w,h,ww,hh,ans,s:integer;function max(p,q:integer):integer;begin if p>q then max:=p else max:=q;end;be 阅读全文

摘要：这题没什么好说的，典型LCS问题，不过注意，poj里的数据，两个字串之间不一定只用空格隔开，读入的时候要注意program poj1458;var f:array[0..1000,0..1000] of integer; s1,s2,s:ansistring; i,j:integer;function max(p,q:integer):integer;begin if p>q then max:=p else max:=q;end;begin while not eof do begin readln(s); s:=s+' '; s1:=''; s2:=& 阅读全文

摘要：求原字串与其反字串的最长公共子序列，最后结果用n减去LCS的长度即可program poj1159;var s1,s2:ansistring; n,i,j:integer; f:array[0..1,0..5001] of integer;function max(p,q:integer):integer;begin if p>q then max:=p else max:=q;end;begin readln(n); readln(s1); fillchar(f,sizeof(f),0); s2:=''; for i:=n downto 1 do s2:=s2+s1[i 阅读全文

摘要：动态规划，其实这题并不难，大概是我dp题做的不够，才没想出动态转移方程，看了discuss里的题解，唉~今后仍需努力啊f[i,j]表示子串s1[1..i]和s2[1..j]的分值。考虑一个f[i,j]，我们有： 1.s1取第i个字母，s2取“-”：f[i-1,j] + score[s1,'-'] 2.s1取“-”，s2取第j个字母：f[i,j-1] + score['-',s2[j]] 3.s1取第i个字母，s2取第j个字母：f[i-1,j-1] + score[s1,s2[j]]即f[i,j] = max(f[i-1,j] + score[s1,'-& 阅读全文

摘要：动态规划，用s[i] 记录前 i 类珍珠的总数。推出状态转移方程：f[i] = min (f[i], (s[i] - s[j] + 10) * price[i] + f[j]);贴个程序：program poj1260;type node=record sum,price:integer; end;var k,m:longint; n,i,j:integer; a:array[1..100] of node; s,f:array[0..100] of longint;function min(p,q:longint):longint;begin if p<q then min:=p el 阅读全文