由于本人水平不够，这场多校只写了第1题和第10题
第1题
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6763
Problem A. Total Eclipse
There are n cities and m bidirectional roads in Byteland. These cities are labeled by 1, 2, . . . , n, the
brightness of the i-th city is bi.
Magician Sunset wants to play a joke on Byteland by making a total eclipse such that the brightness of
every city becomes zero. Sunset can do the following operations for arbitrary number of times:
• Select an integer k (1 ≤ k ≤ n).
• Select k distinct cities c1, c2, . . . , ck (1 ≤ ci ≤ n) such that they are connected with each other. In other words, for every pair of distinct selected cities ci and cj (1 ≤ i < j ≤ k),
if you are at city ci,you can reach city cj without visiting cities not in {c1, c2, . . . , ck}.
• For every selected city ci (1 ≤ i ≤ k), decrease bci by 1.
Now Sunset is wondering what is the minimum number of operations he needs to do, please write a program to help him
Input
The fifirst line of the input contains a single integer T (1 ≤ T ≤ 10), the number of test cases.
For each case, the fifirst line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 200 000),denoting the number of cities and the number of roads.
The second line of the input contains n integers b1, b2, . . . , bn (1 ≤ bi ≤ 109 ), denoting the brightness of each city.
Each of the following m lines contains two integers ui and vi (1 ≤ ui , vi≤ n, ui= vi), denoting an bidirectional road between the ui-th city and the vi-th city. Note that there may be multiple roads
between the same pair of cities.
Output
For each test case, output a single line containing an integer, the minimum number of operations.
题意：给定一个无向图，每个点给定权值，每次选择一个连通块，将连通块里面所有点权值同时减1，直到所有点权值都变成 0，问最少需要几次
解题思路（官方题解）：
每次一定是选择一个极大连通块，将里面所有数同时减小，直到最小值变成 0，然后将变成 0 的点删除，分裂成多个连通块再接着做。
为了实现这个策略，可以将整个过程倒过来看，变成按照 b 的值从大到小依次加入每个点。加入每个点 x 时遍历与 x 相连的所有边 (x, y)，如果 y 在 x 之前加入且 x 和 y 不连通则将 x
和 y 合并，并将 y 所在连通块的树根的父亲设为 x，得到一棵有根树。那么每个点 x 在成为最
小值之前已经被做了 bfatherx 次操作，所以每个点 x 对答案的贡献为 bx-bfatherx。
使用并查集支持路径压缩，时间复杂度 O((n + m)log n)。
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int N=1e5+5; 4 struct node 5 { 6 int id,bi; 7 }ss[N]; 8 int f[N],fa[N],b[N]; 9 bool vis[N]; 10 vector<int > vec[N]; 11 int n,m; 12 int find(int x) 13 { 14 if(f[x]==x)return x; 15 return f[x]=find(f[x]); 16 } 17 void init() 18 { 19 memset(vis,0,sizeof(vis)); 20 memset(fa,0,sizeof(fa)); 21 for(int i=0;i<N;i++) 22 f[i]=i,vec[i].clear(); 23 } 24 bool cmp(struct node x,struct node y) 25 { 26 return x.bi>y.bi; 27 } 28 29 int main() 30 { 31 int t; 32 scanf("%d",&t); 33 while(t--) 34 { 35 init(); 36 scanf("%d%d",&n,&m); 37 for(int i=1;i<=n;i++) 38 { 39 scanf("%d",&b[i]); 40 ss[i].bi=b[i]; 41 ss[i].id=i; 42 } 43 sort(ss+1,ss+1+n,cmp); 44 int u,v,father; 45 for(int i=0;i<m;i++) 46 { 47 scanf("%d%d",&u,&v); 48 vec[u].push_back(v); 49 vec[v].push_back(u); 50 } 51 for(int i=1;i<=n;i++) 52 { 53 u=ss[i].id; 54 vis[u]=1; 55 int len=vec[u].size(); 56 for(int j=0;j<len;j++) 57 { 58 v=vec[u][j]; 59 if(vis[v]==0)continue; 60 father=find(v); 61 if(father==u)continue; 62 fa[father]=f[father]=u; 63 } 64 } 65 long long ans=0; 66 for(int i=1;i<=n;i++) 67 ans+=b[i]-b[fa[i]]; 68 printf("%lld\n",ans); 69 } 70 return 0; 71 }
第10题
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6763
Problem J. Lead of Wisdom
In an online game, “Lead of Wisdom” is a place where the lucky player can randomly get powerful items.
There are k types of items, a player can wear at most one item for each type. For the i-th item, it has
four attributes ai , bi , ci and di . Assume the set of items that the player wearing is S, the damage rate of
the player DMG can be calculated by the formula:
Little Q has got n items from “Lead of Wisdom”, please write a program to help him select which items to wear such that the value of DMG is maximized.
Input
The fifirst line of the input contains a single integer T (1 ≤ T ≤ 10), the number of test cases.
For each case, the fifirst line of the input contains two integers n and k (1 ≤ n, k ≤ 50), denoting the number of items and the number of item types.
Each of the following n lines contains fifive integers ti , ai , bi , ci and di (1 ≤ ti ≤ k, 0 ≤ ai , bi , ci , di ≤ 100),
denoting an item of type ti whose attributes are ai , bi , ci and di.
Output
For each test case, output a single line containing an integer, the maximum value of DMG.
题意：每组选一个，使DMG最大
题解：暴力
我的代码写的太丑了，这里就不给出了
官方题解代码：
1 #include<cstdio> 2 typedef long long ll; 3 const int N=55; 4 int Case,n,m,i,j,x,cnt[N],nxt[N],e[N][N][4];ll ans; 5 void dfs(int x,int a,int b,int c,int d){ 6 if(x>m){ 7 ll tmp=1LL*a*b*c*d; 8 if(tmp>ans)ans=tmp; 9 return; 10 } 11 int num=cnt[x]; 12 if(!num){ 13 dfs(nxt[x],a,b,c,d); 14 return; 15 } 16 for(int i=1;i<=num;i++)dfs(x+1,a+e[x][i][0],b+e[x][i][1],c+e[x][i][2],d+e[x][i][3]); 17 } 18 int main(){ 19 scanf("%d",&Case); 20 while(Case--){ 21 scanf("%d%d",&n,&m); 22 for(i=1;i<=m;i++)cnt[i]=0; 23 while(n--){ 24 scanf("%d",&x); 25 cnt[x]++; 26 for(j=0;j<4;j++)scanf("%d",&e[x][cnt[x]][j]); 27 } 28 x=m+1; 29 for(i=m;i;i--){ 30 nxt[i]=x; 31 if(cnt[i])x=i; 32 } 33 ans=0; 34 dfs(1,100,100,100,100); 35 printf("%lld\n",ans); 36 } 37 }