动态规划转移方程（Dynamic programming transfer equation）

1、动态规划转移方程（Dynamic programming transfer equation）- we make life a weird phase. We put this world into a rich state1. resource issues 1Machine allocation problemFI, j: =max (fi-1, k+wi, j-k)2. resource issues 2-01 knapsack problemFI, j: =max (fi-1, j-vi+wi, fi-1, j);3. linear dynamic programming 1- sim

2、plest longest subsequenceFi: =maxfj+14. subdivision problem 1Stone mergeFi, j: =min (fi, k+fk+1, j+sumi, j);5. subdivision problem 2Polygon subdivisionFI, j: =min (fi, k+fk, j+ak*aj*ai);6. subdivision problem 3Product maximumFi, j: =max (fk, j-1*multk, i);7. resource issues 3System reliability (full

3、 backpack)Fi, j: =maxfi-1, j-ci*k*PI, x8. greedy dynamic programming 1Fast food problemFi, j says the former I line produces J hamburgers, and the most drinks produced by K French fries,The most packages are ans:=minj, div, a, K, div, B, fI, J, k, div, cFi, J, k: =maxfi-1, j, k+ (Ti- (j-j), *p1- (k-

4、k) *p2) div p3Time complexity O (10*1004)9. greedy dynamic programming 2Fi=minf (i-k) stonei (not)f (i-k)+1 (stonei) + greedy compression state10. subdivision problem 4Polygon - Discussion of dynamic programmingFi, j: =maxis positive, fI, k*fk+1, j;A gI, k*fk+1, j;Positive and negative gI, k*fk+1, j

5、;Negative positive fI, k*gk+1, j; G is min11. tree dynamic programming 1Add two trees (from the two sides to the root node model)FI, j: =maxfI, k-1*fk+1, j+ck12. tree dynamic programming 2 selecting courses (multi tree, two tree, top down model)FI, j means the maximum credit obtained by using I as t

6、he root node to select the j gateFi, j: =maxfti.l, k+fti.r, j-k-1+ci13. counting problems 1Weighing weightsConst w:array1.N, of, shortint= (1, 2, 3, 5, 10, 20);/ / different weightsVaR, a:array, 1.N, of, integer;A number of different weight / /F0: =1; total weight (Ans)F1: =0; first weight 0;Ff0+1=f

7、j+k*wj;(1=i=n; 1=j=f0; 1=k=ai;)14., 1- nuclear power plant problemsF-1: =1; f0: =1;Fi: =2*fi-1-fi-1-m15., 2Division of numbersFi, j: =fi-j, j+fi-1, j-1;16. maximum submatrix 1A maximal 01 submatrixFi, j: =min (fi-1, j, vi, j-1, vi-1, j-1) +1;Ans:=maxvalue (f);17. decision questions 1Can it be divisi

8、ble by 4?G1,0: =true; g1,1: =false; g1,2: =false; g1,3: =false;Gi, j: =gi-1, k, and (k+ai, p) mod 4 = J)18. decision questions 2Whether divisible by KFI, j + ni, mod, k: =fi-1, j; -k=j=k; 1=i 0, j 0, x i = y j.max f (i, j) + f i - 1, j (i 0, j 0, x i y j.let (n m), (n = length (a); m: = length (b);f

9、or i: = 1 to n dobeginx: = - 1; p: = 1;for i: = 1 to n doif (a i = b j.beginx: = p;while flag j, x and f j, x (a i) do inc. (x).p: = x;f (j, x): = a i.flag (j, x): = true;endelseif (x - 1) and flag j - 1, x and (not flag j, x) or (f j - 1, x) f (j, x) thenbeginf (j, x): = f j - 1, x;flag (j, x): = t

10、rue;end else x: = - 1;end;ok: = false;for i: = m downto 1 doif flag m then begin writeln (i); right: = true; break; end;if not ok then writeln (0);最大子矩阵2 22.最大带权01子矩阵o - (n 2 * m)枚举行的起始, 压缩进数列, 求最大字段和, 遇0则清零f i: = max (f i-1 + a i, a i)readln (n, m).for i: = 1 to n do for i: = 1 to n do read (a i, j

11、);the: = - maxlongint;for i: = 1 to n dobeginfillchar (b, sizeof (b), 0);fillchar (u, sizeof (u), 0);for i: = 1 to n dobeginmax: = 0;for k = 1 to n dobeginif (a j, k 0) and (not u k) thenbegininc. (b k, a j, k);inc. (max, b k)endelsebeginmax: = 0;u k: = true;end;if max to then the: = max.end;end;end

12、;23. 资源问题4装箱问题 - (判定性01背包)f j: = (f j and f j - v i).注: 这里将数字三角形的意义扩大凡状态转移为图形, 跟其上面阶段和前面状态有关都叫数字三角形:)数字三角形1 24.朴素数字三角形 -f i, j: = max (f i + 1, j + a i, j, f i + 1, j + 1 + a i, j);数字三角形2 25.晴天小猪历险记之hill -同一阶段上暴力动态规划if i,：= min（F 我，我J-1，F，J + 1 ， J ，表示F，F【I-1、J-1 ）+一个我26双向动态规划1。数字三角形3-小胖办证F 我：= max（

13、F【I-1、J + 我，我，F，J 1 + 我，我的F，J + 1 + 我）27。数字三角形4-过河卒/ /边界初始化F 我：= F + F，我，J-1；28数字三角形5。-朴素的打砖块F 我，J，K ：= max（F【I-1，JK，P + 我和我，K，F，J，K ）；29数字三角形6。-优化的打砖块F 我，J，K ：= max 【I-1，JK，K-1 和我+，K 30线性动态规划3。-打鼹鼠”F 我：= F J + 1；（ABS（x 我 X J ）+ ABS（Y 我 - J ） = T 我 - J ）31树形动态规划3。-贪吃的九头龙32状态压缩动态规划1。-炮兵阵地马克斯（f *（r +

14、1）+ K），g（j）如果（map（i）和计划k0）和（计划P或计划问）和计划 k = 0）33递推天地3。-情书抄写员F 我：= F + K * F 的34递推天地4。-错位排列F 我：=（i-1）（F【I-2 + F）；F ：= N * F【n-1】+（- 1）（n-2）；35递推天地5。-直线分平面最大区域数f：= n（n + 1）div 2 + 1；36递推天地6。-折线分平面最大区域数f：=（n-1）（2 * n-1）+ 2 * n；37递推天地7。-封闭曲线分平面最大区域数f：= 2。：SQR（n）n + 2；38递推天地8-凸多边形分三角形方法数F ：= c（2n，n-1）DIV

15、 N；对于K边形F k ：= c（2K-4，K-2）div（k-1）；/ /（K3）39递推天地9-加泰罗尼亚数列一般形式1,1,2,5,14,42132fn（2K，k）div（k + 1）；40递推天地10-彩灯布置排列组合中的环形染色问题F ： 1 * = f（-2）+ F 2 *（m-1）；（F 1 ：= m；F 2 ：= M（M-1）；41线性动态规划4-找数线性扫描求和；（如果金额=目标然后去；如果和目的然后公司（我）其他公司（J）；）42线性动态规划5-隐形的翅膀闵：= min ABS（W / W 我 J -金）；如果W / W的我 J = F 我，Q，P F 我然后打破别人，Q，

16、P ：= xs（x，k）；公司（K）；计数问题2-陨石的秘密（排列组合中的计数问题）和 L1，L2，L3，D： L1 = F + 1，L2、L3，(d + 1) - f (l1 + 1, l2, l3, and d.f (l1, l2, l3, d) = sigma (f (o, p, q, d - 1) * f (l1, l2), (l3 - q).线性动态规划合唱队形 -两次f i: = max s + 1 枚举中央结点 j资源问题明明的预算方案 加花的动态规划 -.f i, j = max (f i, j, f i, j - v i v bf - v i f i + v i p i, +

17、 v * p i. for i) + v (f i p f i).资源问题化工场装箱员.树形动态规划聚会的快乐.(f) 1, 2: = max (f (1, 0), f (i, 1).f (i, 1) = sigma (f t i, economic, 0).f 1, 0 = sigma (f t i, economic, 3).树形动态规划皇宫看守.(f) 1, 2: = max (f (1, 0), f (i, 1).f (i, 1) = sigma (f t i, economic, 0).f 1, 0 = sigma (f t i, economic, 3).递推天地盒子与球.f (i

18、, 1) = 1.f i, j: = j (f i, 1, j - 1) + f (1, j).双重动态规划有限的基因序列.f i: = min s + 1 j.g (c, i, j): = (g (a, i, j) and g (b, i, j), or (g (c, i, j)最大子矩阵问题居住空间.f (i, j, k): = min (min (min (f (i), (j), (k), f (i, j), (k).min (f i, j, k, - 1, f i, 1, j, k).min (min (f (i), (j), (k) - 1, f i, j, k, 1).f (i),

19、 (j), (k - 1) + 1).线性动态规划日程安排 -f i: = max f j + p i; (c j s i)递推天地组合数 -c i, j: = c i - 1, j + c i, 1, j - 1) c i: = 1 0 0 0 0 0 0树形动态规划有向树k中值问题.f (i, r, k): = max max f (s (i), (i), (j) + f r i (i, k, j - 1), (f (f (s i, r, j + f r i. r, k, j) + w (i, r)树形动态规划sctc 2001选课.f i, j: = (w i (if ip) + f (

20、s i, k + f i m - k (0km) (if the i (0)线性动态规划多重历史.f i, j = sigma f (i, k, j - 1) (if checked.背包问题 (+ + 回溯 ceoi1998 substract 1背包问题).f i, j: = f i, 1, j - i or i-1, j + a i)线性动态规划 (字符串)in 2000 古城之谜.f i: = min f (i + length (s), 2.1, f i + length (s) + 1 f (1.1) (1.2): = min s i + length (s) 2 + words

21、s, f i + length (s) 2 + words s.线性动态规划最少单词个数.f i, j = max (f i, j, f n - 1, j - 1) + 1线型动态规划apio2007 数据备份.2个状态和j 状态压缩剪掉每个阶段j前j * * 2 + 200后的状态贪心动态规划f i: = min (g i (2) + s i f - 1).树形动态规划apio2007 风铃.f i: = f t + f r + 1 - c i (c) r.g i: = (s i d (r) 0 (d) i = ()g t = g (r) = 1, then halt.地图动态规划in 20

22、05 adv19910.f (t, (i, j): = max f (t 1, t - d d a, j - dy d k + 1, f t - 1, i, j).地图动态规划优化的noi adv19910), 2005f (k, i, j): = max f (k - 1, i, p) + 1 j - b k = p = (j);目标动态规划ceoi98 subtra.f i, j: = f i, 1, j + a i or i-1, j - 1.目标动态规划vijos 1037搭建双塔问题.e delta: = g value (value + a i, delta + (i) and g

23、 value, delta - (i)树形动态规划有线电视网.f (i,P ：= max（F 我，P，F 我，P-Q + F J，Q 图我）叶我 = P = l，1 = Nb is marked 0, otherwise the standard 1 is compressed into the K in binary state, in which case the minimum cost is spentFI, J, k, =minfl, u, K, and (si (i-1)+w1, fr, j-u, K, and (si (i-1)Tree dynamic programming-I

24、OI2005 RiverFi: =maxMemory search-Vijos something, I forgotFpre, h, m: =sigmaSDP (I, h+1, M+i) (Pre=i=M1)State compression dynamic programming-APIO 2007 ZooFI, k: =fi-1, K, and, not (1=0, then, Min (fi, j, fi-1, j-k+time (I, K);knapsack problem- USACO Raucous RockersMultiple backpacks cannot be plac

25、ed repeatedly, but the order of items is limited.FI, J, and k represent decisions to I items, J packs, which take up K of space.FI, J, k: =max (fI-1, J, k, fI-1, J, J-1, k-ti+pi, fi-1, maxtime-ti)Multi process dynamic programmingCruise Canada (IOI95, USACO)Di, j=maxdk, j+1 (ak, i & jki), dj, k+1 (aI

26、, j & (kjThe analysis state (I, J), it may be (k, J) (jki) that K arrives at I to get (mode 1), or that it is (J, K) (kj) that K exceeds J and arrives at I (mode 2). But it cant be (I, K) (kj) that K arrives at J because it might have duplicate paths. Even if there is no duplicate path, then it can

27、be obtained by the same way (J, K), so that the time complexity of O is not missing (N3) 2dynamic programming-ZOJ cheeseFi, j: =fi-kk*zlu, 1, j-kk*zlu, 2+ai-kk*zlu, 1, j-kk*zlu, 2dynamic programming-NOI 2004 berry linearFI, 1: =siFI, j: =maxminsi-sl-1, fl-1, j-1 (2 = J = k, j = L = I)dynamic programming-NOI 2004 berry complete undirected graphFI, j: =fi-1, j or (J = wi) and (fi-1, j-wi)dynamic programmingGravel combined quadrilater