全球网站流量排名查询ceo是什么职位

全球网站流量排名查询,ceo是什么职位,免费查企业信息查询,king cms网站建设题目描述 Petra\texttt{Petra}Petra 和 Jan\texttt{Jan}Jan 收到 nnn 个礼物#xff0c;每个礼物对 Petra\texttt{Petra}Petra 的价值为 pip_ipi​ #xff0c;对 Jan\texttt{Jan}Jan 的价值为 jij_iji​ 。两人轮流挑选礼物#xff0c;通过抛硬币决定谁先开始。 Petra\text…题目描述Petra\texttt{Petra}Petra和Jan\texttt{Jan}Jan收到nnn个礼物每个礼物对Petra\texttt{Petra}Petra的价值为pip_ipi​对Jan\texttt{Jan}Jan的价值为jij_iji​。两人轮流挑选礼物通过抛硬币决定谁先开始。Petra\texttt{Petra}Petra采用贪心策略每次选择对自己价值pip_ipi​最高的礼物如果有多个礼物pip_ipi​相同则选择对Jan\texttt{Jan}Jan价值jij_iji​最低的礼物。Jan\texttt{Jan}Jan的策略则是最优化自己的总价值并且在多个选择获得相同价值时让Petra\texttt{Petra}Petra的总价值尽可能大。给定初始的抛硬币结果谁先手以及每个礼物对两人的价值求最终两人各自获得的总价值按各自的评估标准。输入格式第一行测试用例数量TTT最多100100100个。每个测试用例一行整数nnn1≤n≤10001 \le n \le 10001≤n≤1000礼物数量。一行字符串Petra\texttt{Petra}Petra或Jan\texttt{Jan}Jan表示先手的人。nnn行每行两个整数pip_ipi​和jij_iji​0≤pi,ji≤10000 \le p_i, j_i \le 10000≤pi​,ji​≤1000分别表示Petra\texttt{Petra}Petra和Jan\texttt{Jan}Jan对第iii个礼物的评估价值。输出格式每个测试用例输出一行包含两个整数Petra\texttt{Petra}Petra获得的总价值和Jan\texttt{Jan}Jan获得的总价值。题目分析本题的关键在于理解两人的策略差异并设计合适的状态转移方程。1. 策略分析Petra\texttt{Petra}Petra的贪心策略她的选择顺序是固定的总是选择当前剩余礼物中对自己价值最高ppp最大的礼物ppp相同时选对Jan\texttt{Jan}Jan价值最低jjj最小的礼物。因此我们可以将所有礼物按照Petra\texttt{Petra}Petra的优先级排序ppp降序ppp相同时jjj升序。排序后Petra\texttt{Petra}Petra总是从前往后依次选取礼物。Jan\texttt{Jan}Jan的最优策略Jan\texttt{Jan}Jan知道Petra\texttt{Petra}Petra的选取顺序他可以在自己的回合中选择任意一个礼物目标是在游戏结束时最大化自己的总价值并且在多个最优方案中选择让Petra\texttt{Petra}Petra总价值最大的那个。2. 问题转化由于Petra\texttt{Petra}Petra的选取顺序固定问题可以转化为在排序后的礼物列表中Jan\texttt{Jan}Jan可以选择“抢占”某些Petra\texttt{Petra}Petra将要拿的礼物。具体来说每个礼物有两个状态被Jan\texttt{Jan}Jan拿走或被Petra\texttt{Petra}Petra拿走。Jan\texttt{Jan}Jan的决策可以看作对于排序后的第iii个礼物Jan\texttt{Jan}Jan可以选择是否抢占它。如果抢占Jan\texttt{Jan}Jan获得jij_iji​价值Petra\texttt{Petra}Petra失去pip_ipi​价值因为她拿不到这个礼物了如果不抢占Petra\texttt{Petra}Petra正常拿走该礼物获得pip_ipi​价值。3. 动态规划设计定义状态dp[i][k]dp[i][k]dp[i][k]考虑前iii个礼物Jan\texttt{Jan}Jan抢占了kkk个时Jan\texttt{Jan}Jan能获得的最大总价值。val[i][k]val[i][k]val[i][k]在上述情况下Jan\texttt{Jan}Jan抢占的礼物的Petra\texttt{Petra}Petra价值之和即Petra\texttt{Petra}Petra因此损失的价值。状态转移对于第iii个礼物排序后Jan\texttt{Jan}Jan不抢占Petra\texttt{Petra}Petra拿走该礼物。此时Jan\texttt{Jan}Jan的价值不变Petra\texttt{Petra}Petra的损失不变。dp[i][k]dp[i−1][k]dp[i][k] dp[i-1][k]dp[i][k]dp[i−1][k]val[i][k]val[i−1][k]val[i][k] val[i-1][k]val[i][k]val[i−1][k]Jan\texttt{Jan}Jan抢占Jan\texttt{Jan}Jan拿走该礼物。此时Jan\texttt{Jan}Jan的价值增加jij_iji​Petra\texttt{Petra}Petra的损失增加pip_ipi​。dp[i][k]dp[i−1][k−1]jidp[i][k] dp[i-1][k-1] j_idp[i][k]dp[i−1][k−1]ji​val[i][k]val[i−1][k−1]pival[i][k] val[i-1][k-1] p_ival[i][k]val[i−1][k−1]pi​决策时优先最大化dp[i][k]dp[i][k]dp[i][k]Jan\texttt{Jan}Jan的价值如果dp[i][k]dp[i][k]dp[i][k]相同则选择val[i][k]val[i][k]val[i][k]较小的方案让Petra\texttt{Petra}Petra损失更小即Petra\texttt{Petra}Petra最终价值更大。4. 先手影响Jan\texttt{Jan}Jan最多能抢占的礼物数量取决于谁先手Petra\texttt{Petra}Petra先手两人轮流Jan\texttt{Jan}Jan最多能抢占⌊n/2⌋\lfloor n/2 \rfloor⌊n/2⌋个礼物。Jan\texttt{Jan}Jan先手Jan\texttt{Jan}Jan可能多抢一个最多能抢占⌈n/2⌉\lceil n/2 \rceil⌈n/2⌉个礼物。在动态规划过程中对于前iii个礼物Jan\texttt{Jan}Jan最多能抢占的数量numnumnum为Petra\texttt{Petra}Petra先手numi/2num i/2numi/2Jan\texttt{Jan}Jan先手num(i1)/2num (i1)/2num(i1)/25. 结果计算设所有礼物的Petra\texttt{Petra}Petra价值总和为totalPtotalPtotalP。在动态规划结束后我们遍历所有可能的kkkJan\texttt{Jan}Jan抢占的数量找到dp[n][k]dp[n][k]dp[n][k]最大的方案如果多个方案dp[n][k]dp[n][k]dp[n][k]相同选择val[n][k]val[n][k]val[n][k]最小的。最终Jan\texttt{Jan}Jan的总价值 dp[n][k]dp[n][k]dp[n][k]Petra\texttt{Petra}Petra的总价值 totalP−val[n][k]totalP - val[n][k]totalP−val[n][k]时间复杂度排序复杂度O(nlog⁡n)O(n \log n)O(nlogn)动态规划状态数O(n2)O(n^2)O(n2)总时间复杂度O(n2)O(n^2)O(n2)在n≤1000n \le 1000n≤1000时可行。代码实现// Free Goodies// UVa ID: 12260// Verdict: Accepted// Submission Date: 2025-12-10// UVa Run Time: 0.120s//// 版权所有C2025邱秋。metaphysis # yeah dot net#includebits/stdc.husingnamespacestd;structCandy{intp,j;};boolcmp(constCandya,constCandyb){if(a.p!b.p)returna.pb.p;returna.jb.j;}intmain(){intt;cint;while(t--){intn;cinn;string first;cinfirst;vectorCandycandies(n);inttotalP0;for(inti0;in;i){cincandies[i].pcandies[i].j;totalPcandies[i].p;}sort(candies.begin(),candies.end(),cmp);vectorvectorintdp(n1,vectorint(n1,0));vectorvectorintval(n1,vectorint(n1,0));for(inti1;in;i){intnum;if(firstPetra){numi/2;}else{num(i1)/2;}for(intj1;jnum;j){dp[i][j]dp[i-1][j];val[i][j]val[i-1][j];intcandJandp[i-1][j-1]candies[i-1].j;intcandValval[i-1][j-1]candies[i-1].p;if(candJandp[i][j]){dp[i][j]candJan;val[i][j]candVal;}elseif(candJandp[i][j]candValval[i][j]){val[i][j]candVal;}}}intmaxJan0,minVal0;intmaxTake;if(firstPetra){maxTaken/2;}else{maxTake(n1)/2;}for(intj1;jmaxTake;j){if(dp[n][j]maxJan){maxJandp[n][j];minValval[n][j];}}intpetraValtotalP-minVal;coutpetraVal maxJanendl;}return0;}