POJ 3723 Conscription——————求森林最大权

D4与D5平台对显卡性能的真实影响:PCIe带宽、内存延迟与DMI瓶颈深度解析 PCIe带宽和内存子系统是决定GPU实际性能的关键底层因素,而非仅由显卡自身参数决定。PCIe协议版本(如4.0/5.0)、通道拓扑结构、CPU直连能力,以及DDR4/DDR5在延迟与带宽上的差异,共同构成数据通路的‘吞吐—延迟—调度’三角约束。其中,DMI总线带宽限制常被忽视,却在直播、AI推理、多任务渲染等混合IO场景中成为性能拐点;而Resizable BAR与PCIe链路训练质量则直接影响GPU能否稳定吃满系统资源。本文基于实测数据,揭示D4/D5平台在真实工作负载下的性能分水岭,为AI训练、实时渲 阅读详情

Conscription

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 19544 Accepted: 6794
Description

Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl x and boy y have a relationship d and one of them has been collected, Windy can collect the other one with 10000-d RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

Input

The first line of input is the number of test case.
The first line of each test case contains three integers, N, M and R.
Then R lines followed, each contains three integers xi, yi and di.
There is a blank line before each test case.

1 ≤ N, M ≤ 10000
0 ≤ R ≤ 50,000
0 ≤ xi < N
0 ≤ yi < M
0 < di < 10000

Output

For each test case output the answer in a single line.
Sample Input

2

5 5 8
4 3 6831
1 3 4583
0 0 6592
0 1 3063
3 3 4975
1 3 2049
4 2 2104
2 2 781

5 5 10
2 4 9820
3 2 6236
3 1 8864
2 4 8326
2 0 5156
2 0 1463
4 1 2439
0 4 4373
3 4 8889
2 4 3133
Sample Output

71071
54223


中文意思
在这里插入图片描述

可以转化为无向图中的最大权森林问题
可以把变得权值变成负的然后可以用最小生成树算法求解

// #include<bits/stdc++.h>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
const int MAXN = 5e4+5;
int p[MAXN];
struct edge{
        int from,to,cost;
        edge(){};
        edge(int _from, int _to, int _cost){
                from = _from;
                to = _to;
                cost = _cost;
        }
}es[MAXN];

void init(int n){
        for(int i=0;i<=n;++i){
                p[i] = i;
         }
}
int find(int x){
        return p[x]==x?x:p[x]=find(p[x]);
}
void join(int x, int y){
        x = find(x);
        y = find(y);
        if(x!=y)        p[y] = x;
}

bool cmp(edge x, edge y){
        return x.cost<y.cost;
}


int main()
{
        // freopen("../in.txt","r",stdin);
        // freopen("../out.txt","w",stdout);
        int t,n,m,r,x,y,d;
        scanf("%d",&t);
        while(t--){
                scanf("%d %d %d",&n,&m,&r);
                init(n+m);
                for(int i=0;i<r;++i){
                        scanf("%d %d %d",&x,&y,&d);
                        es[i] = edge(x,n+y,-d);
                }
                sort(es,es+r,cmp);
                int sum = 0;
                for(int i=0;i<r;++i){
                        edge e = es[i];
                        if(find(e.from)!=find(e.to)){
                                sum += e.cost;
                                join(e.from, e.to);
                        }
                }
                printf("%d\n",10000*(n+m)+sum);

        }
        return 0;
}

《Ventoy 制作系统安装 U 盘的具体步骤》 然后您可以在电脑的磁盘管理中对 U 盘进行快速格式化,选择 FAT32 或 NTFS 文件系统。至此,Ventoy 系统安装 U 盘就制作完成了。您可以在需要安装系统的电脑上,从 U 盘启动,然后选择相应的系统镜像进行安装。安装完成后,将您需要的操作系统镜像文件(如 Windows、Linux 等的 ISO 格式镜像)复制到 U 盘的根目录。在安装程序中,选择您插入的 U 盘设备,确保选择正确,以免误操作。2. 确认选择的 U 盘盘符正确,防止误操作导致其他磁盘数据丢失。点击“安装”按钮,等待安装完成。 阅读详情

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plot(cnn.rL);assert(er<0.12, 'Too big error');```Example: Neural Networks---------------------```matlabfunction test_example_NNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);% normalize[train_x, mu, sigma] = zscore(train_x);test_x = normalize(test_x, mu, sigma);%% ex1 vanilla neural netrand('state',0)nn = nnsetup([784 100 10]);opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samples[nn, L] = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.08, 'Too big error');%% ex2 neural net with L2 weight decayrand('state',0)nn = nnsetup([784 100 10]);nn.weightPenaltyL2 = 1e-4; % L2 weight decayopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex3 neural net with dropoutrand('state',0)nn = nnsetup([784 100 10]);nn.dropoutFraction = 0.5; % Dropout fraction opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex4 neural net with sigmoid activation functionrand('state',0)nn = nnsetup([784 100 10]);nn.activation_function = 'sigm'; % Sigmoid activation functionnn.learningRate = 1; % Sigm require a lower learning rateopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex5 plotting functionalityrand('state',0)nn = nnsetup([784 20 10]);opts.numepochs = 5; % Number of full sweeps through datann.output = 'softmax'; % use softmax outputopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex6 neural net with sigmoid activation and plotting of validation and training error% split training data into training and validation datavx = train_x(1:10000,:);tx = train_x(10001:end,:);vy = train_y(1:10000,:);ty = train_y(10001:end,:);rand('state',0)nn = nnsetup([784 20 10]); nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

最大权森林·POJ-3723·Conscription

题目大意:就是要找n个女兵,m个男兵,每招一个兵的价钱是10000;然后给出男女兵之间的亲密度。亲密度的作用就是用来省钱,就是要找一个兵时,在已招的兵中,有存在与这个兵有亲密度的,这个兵的价格就为10000-max(亲密度)(就是与已招的兵中最大亲密度);思路:这边其实最终是要求最大权值森林,然后这边讲一下为什么可以转化成最大权值森林,以及怎么求解最大权值森林。1.为什么可以转化成最大权值森林?首...

吾之奉先的博客 387

POJ 3723 Conscription 最小生成树

【代码】POJ 3723 Conscription 最小生成树。

qq_53038151的博客 133

POJ 3723 Conscription (最小生成树)

题目链接 题目大意: 解题思路: 本题最大干扰是男女配对,其实这个条件一点用都没有,不要想复杂了(想成了二分图,我一开始就是这么想的。。。。) 题目要求招募n个女兵,m个男兵;但是一共就n个女兵,m个男兵,也就是说全都招募了 解法很明显,让(10000 - di)求和最小,也就是边权di求和最大,就是“最大生成树“; 那么就来实现一下最大生成树算法就行! 最大生成树算法(其实就是最小生成树的变种) 把表达式变形一下就是 10000 * (n + m) + (-d1 - d2 - d3 - d4 -

364

POJ 3723(最小生成树+优先队列)

POJ 3723 原题链接:http://poj.org/problem?id=3723 大致题意:招募n名男生与m名女生,没招募一位,底薪是10000RMB,若i男与j女之间存在亲密值,那么实际支付10000-k。 求招募n+m员工 所花最小费用 思路:最小生成树无疑了,将男生的编号加上m或者将女生编号加上n,那么就让女生和男生称为一体,换句话说 顶点数为n+m。 求最小生成树有两种方法:1.prime算法。2.kruskal算法。 本题可用kruskal(没试过prime)+并查集维护边+优先队列(so

weixin_51596558的博客 556

POJ 3723(最大生成树+kuruakal)

Conscription Time Limit: 1000MS Memory Limit: 65536K Description Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect...

诗意的栖居 503

poj3723--Conscription(最大权森林问题)

题意:招募n女,m男,其中r对有亲密关系,招募每个人花费为10000-(已经招募的人中和自己亲密度的最大值),求招募所有人的最小总花费。 思路:乍一看以为是二分图,吓得想了一下觉得还是二分图,但其实这个思路转化一下就是最大生成树模板题;可以看作先把所有人招募进来之后减掉亲密关系的优惠。亲密关系的最大优惠即为最大生成树的结果的权值。 错误思路:我的第一个思路是用最小生成树加上一个sum数组记录根节点...

weixin_44756457的博客 255

POJ3723 Conscription 【并检查集合】

Conscription Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8071   Accepted: 2810 Description Windy has a country, and he wants to build an army t...

weixin_34279184的博客 149

POJ3723

Description 链接 Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without...

咸鱼 490

poj3723

题意:国王想n个女孩和m个男孩去当兵,买一个人得花10000块钱。有一个机会可以省钱,题目给出了R种关系,一个男孩和一个女孩有关系,如果他们其中一个人已经被买了,买另一个人就可以少花一些钱。 题解:可以把n个女孩和m个男孩当做图上的点,选所有的点就是使所有点连通,边权是花费,没有关系的两个人花费就是10000,有关系的就是10000-d。 然后通过关系建图就好,主要是理解了题然后最小生成树。

constbh的博客 357

POJ 3723 解题报告

这道题是最小生成树问题。由于“收集”每个人只能用一次“关系”,所以利用的关系不能形成环。贪心从最小的关系开始,只要不能形成环就收集。这就是kruskal用并查集的算法。最后没收集的人每人按最大代价加入总代价就可以了。 thestoryofsnow 3723 Accepted 908K 344MS C++ 2132B /* ID: thestor1 L

thestoryofsnow的专栏 572
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