水题：代码例如以下
#include <bits/stdc++.h>
#define foreach(it,v) for(__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)
using namespace std;
const int maxn = 2e5;
int a[maxn];
int main(int argc, char const *argv[])
{
ios_base::sync_with_stdio(false);
int n,w[3][2];
while(cin>>n) {
for(int i = 0; i < 3; i++) cin>>w[i][0]>>w[i][1];
int a[3];
int limt = w[1][0] + w[2][0];
a[0] = min(w[0][1],n - limt);
n -= a[0];
limt = w[2][0];
a[1] = min(w[1][1],n - limt);
n -= a[1];
a[2] = min(n,w[2][1]);
printf("%d %d %d
", a[0],a[1],a[2]);
}
return 0;
}
#include <bits/stdc++.h>
#define foreach(it,v) for(__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)
using namespace std;
const int maxn = 2e5;
int a[maxn];
int main(int argc, char const *argv[])
{
ios_base::sync_with_stdio(false);
int n,w;
while(cin>>n>>w) {
for(int i = 1; i <= n+n; i++) {
cin>>a[i];
}
sort(a+1,a+1+n+n);
double M = a[n + 1] / 2.0;
M = min(M,a[1]*1.0);
double ans = M * n + 2*M*n;
if(ans > w) ans = w;
printf("%.10f
", ans);
}
return 0;
}枚举最高的那个脚的高度H。大于H的直接删掉(后缀和记录代价即可),设高度为H的有x根,小于H的有y根。那么我们必须删掉k = max(y-(x-1),0)根，本题d值较小开个cnt数组记录下即可，假设比較大的话能够用平衡树维护前k小的代价和。
#include <bits/stdc++.h>
#define foreach(it,v) for(__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)
using namespace std;
const int maxn = 2e5;
int l[maxn],d[maxn];
vector<int> g[maxn];
int s[maxn];
int main(int argc, char const *argv[])
{
ios_base::sync_with_stdio(false);
int n;
while(cin>>n) {
for(int i = 0; i <= 100001; i++) {
s[i] = 0;
g[i].clear();
}
for(int i = 1; i <= n; i++) cin>>l[i];
for(int i = 1; i <= n; i++) {
int x;cin>>x;
g[l[i]].push_back(x);
s[l[i]] +=x;
}
vector<int> t;
for(int i = 100000; i >= 0; i--) s[i] += s[i+1];
for(int i = 0; i <= 100000; i++)if(g[i].size()){
t.push_back(i);
}
int cost = 1e8,sum = 0;
int cnt[205];
memset(cnt,0,sizeof cnt);
foreach(id,t) {
int & x = *id;
int w = s[x+1] - s[100001];
int sz = g[x].size();
int d = sz - 1;
d = max(sum-d,0);
int i;
for(i = 0; i <= 200; i++) {
d -= cnt[i];
w += i*cnt[i];
if(d <= 0) break;
}
w += d*i;
cost = min(cost,w);
sum += g[x].size();
foreach(it,g[x]) {
++cnt[*it];
}
}
printf("%d
", cost);
}
return 0;
}After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
The first line of the input contains two integers n and m (
— the
number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi(1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
4 4 1 2 1 3 4 2 4 3
1 2
3 3 1 2 2 3 3 1
0 1
3 0
3 1
The simple cycle is a cycle that doesn't contain any vertex twice.
首先假设有0条边那么任选3个点就能构成奇环。答案是3 c(n,3)
假设每一个节点最多仅仅有一条边与其关联且m > 0,那么对于每条边(u,v),任选节点k。仅仅须要加入两条边就能构成奇环。答案是2,m*(n-2)。
假设存在节点的度大于2,那么推断是否是否是二分图。假设不是意味着已经存在奇环，答案是 0，1
否则对每一个联通分量仅仅须要在同色节点里任选两个加边就能形成奇环，答案是1,,sum(c(ci[0],2)+c(ci[1],2)）；
盗用Quailty的代码QAQ......
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
vector<int>e[100005];
int vis[100005],cnt[2];
bool bfs(int st)
{
queue<int>q;
vis[st]=1;
q.push(st);
while(!q.empty())
{
int u=q.front();
q.pop();
cnt[vis[u]-1]++;
for(int i=0;i<e[u].size();i++)
{
if(!vis[e[u][i]])
{
vis[e[u][i]]=3-vis[u];
q.push(e[u][i]);
}
else
{
if(vis[e[u][i]]==vis[u])return 0;
}
}
}
return 1;
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
int a,b;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
e[a].push_back(b);
e[b].push_back(a);
}
int maxd=0;
for(int i=1;i<=n;i++)
{
maxd=max(maxd,(int)e[i].size());
}
if(maxd==0)
{
printf("3 %I64d
",1LL*n*(n-1)*(n-2)/6);
}
else if(maxd==1)
{
printf("2 %I64d
",1LL*m*(n-2));
}
else
{
bool odd=0;
ll ans=0;
for(int i=1;i<=n;i++)
{
if(!vis[i])
{
cnt[0]=cnt[1]=0;
if(!bfs(i))
{
odd=1;
break;
}
ans+=1LL*cnt[0]*(cnt[0]-1)/2;
ans+=1LL*cnt[1]*(cnt[1]-1)/2;
}
}
if(odd)printf("0 1
");
else printf("1 %I64d
",ans);
}
return 0;
}Tomorrow Ann takes the hardest exam of programming where she should get an excellent mark.
On the last theoretical class the teacher introduced the notion of a half-palindrome.
String t is a half-palindrome, if for all the odd positions i (
)
the following condition is held: ti = t|t| - i + 1,
where |t| is the length of string t if
positions are indexed from 1. For example, strings "abaa",
"a", "bb", "abbbaa"
are half-palindromes and strings "ab", "bba"
and "aaabaa" are not.
Ann knows that on the exam she will get string s, consisting only of letters a and b, and number k. To get an excellent mark she has to find the k-th in the lexicographical order string among all substrings of s that are half-palyndromes. Note that each substring in this order is considered as many times as many times it occurs in s.
The teachers guarantees that the given number k doesn't exceed the number of substrings of the given string that are half-palindromes.
Can you cope with this problem?
The first line of the input contains string s (1 ≤ |s| ≤ 5000), consisting only of characters 'a' and 'b', where |s| is the length of string s.
The second line contains a positive integer k — the lexicographical number of the requested string among all the half-palindrome substrings of the given string s. The strings are numbered starting from one.
It is guaranteed that number k doesn't exceed the number of substrings of the given string that are half-palindromes.
Print a substring of the given string that is the k-th in the lexicographical order of all substrings of the given string that are half-palindromes.
abbabaab 7
abaa
aaaaa 10
aaa
bbaabb 13
bbaabb
By definition, string a = a1a2... an is lexicographically less than string b = b1b2... bm, if either a is a prefix of b and doesn't coincide with b, or there exists such i, that a1 = b1, a2 = b2, ... ai - 1 = bi - 1, ai < bi.
In the first sample half-palindrome substrings are the following strings — a, a, a, a, aa, aba, abaa, abba, abbabaa, b, b, b, b, baab,bab, bb, bbab, bbabaab (the list is given in the lexicographical order).
半回文串指的是对于每一个奇数位置i(1=<i<=(L+1)/2),满足s[i] = s[L-i+1],下标从1開始，给出一个字符串，求字典序第k小的半回文子串。
Trie的应用，挺奇妙的，把每一个回文子串插到Trie中去。cnt[u]记录u这个子树有多少个串。然后在树上查找即可，
不能暴力插子串。能够用一个数组ok[i][j]表示子串s[i...j]是否是半回文串,然后对每一个后缀suffix(x)插入。假设ok[x][i]为true就更新cnt数组,插入完之后须要dfs一遍求子树的siz。查找算法例如以下，假设当前到了节点root，子树1和2的大小分别为w1和w2,那么显然以u为结束节点的子串有w = cnt[root]-w1-w2个，假设w>=k那么算法终止。否则更新k = k - w。假设w1 >= k，那么进入子树1，否则更新k = k-w1，进入子树2。
#include <bits/stdc++.h>
#define foreach(it,v) for(__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)
using namespace std;
const int maxn = 5e3 + 10;
typedef long long ll;
char s[maxn];
bool ok[maxn][maxn];
int ch[maxn*maxn][2],cnt[maxn*maxn];
struct Trie
{
int tot;
Trie() {
cnt[tot = 0] = 0;
ch[0][0] = ch[0][1] = -1;
}
void Insert(const char *s,int st) {
int u = 0;
for(int i = st; s[i]; i++) {
int c = s[i] - 'a';
if(ch[u][c]==-1) {
ch[u][c] = ++tot;
cnt[tot] = 0;
ch[tot][0] = ch[tot][1] = -1;
}
u = ch[u][c];
if(ok[st][i])++cnt[u];
}
}
int Init(int u) {
if(u==-1) return 0;
cnt[u] += Init(ch[u][0]);
cnt[u] += Init(ch[u][1]);
return cnt[u];
}
};
int pre(char * s)
{
s[0] = ' ';
int n = strlen(s);
--n;
for(int w = 1; w <= n; w++) {
for(int x = 1; x + w -1 <= n; x++) {
ok[x][x+w-1] = (s[x]==s[x+w-1]);
if(w >= 5) ok[x][x+w-1] &= ok[x+2][x+w-3];
}
}
return n;
}
void solve(int root,int k)
{
if(k < 1||root==-1) return;
int v1 = ch[root][0],v2 = ch[root][1];
int w1= (v1 == -1 ? 0 : cnt[v1]),w2 = (v2 == -1? 0: cnt[v2]);
int w = cnt[root] - w1 - w2;
if(w >= k) return;
k -= w;
if(w1 >= k) {
putchar('a');
solve(v1,k);
}else {
k -= w1;
putchar('b');
solve(v2,k);
}
}
int main()
{
// ios_base::sync_with_stdio(false);
// cin.tie(0);
int k;
while(scanf("%s%d",s+1,&k)==2) {
int n = pre(s);
Trie A;
for(int i = 1; i <= n; i++) A.Insert(s,i);
A.Init(0);
solve(0,k);
putchar('
');
}
return 0;
}