Codeforces Round #340 (Div. 2) （629A，629B，629C（排列组合，动态规划），629D（线段树））

Far Relative’s Birthday Cake

题目链接：

http://www.codeforces.com/problemset/problem/629/A

解题思路：

Consider that we have rowi chocolates
in the i'th row and coli chocolates
in the i'th column.

The answer to the problem would be: 

.
It is obvious that every pair would be calculated exactly once (as we have no more than one chocolate in the same square)

Time Complexity: O(n2)

AC代码：

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

char a[110][110];

int main(){
int n;
while(~scanf("%d",&n)){
int ans = 0,tmp;
for(int i = 0; i < n; i++){
scanf("%s",a[i]);
tmp = 0;
for(int j = 0; j < n; j++){
if(a[i][j] == 'C')
tmp++;
}
ans += tmp*(tmp-1)/2;
}
for(int i = 0; i < n; i++){
tmp = 0;
for(int j = 0; j < n; j++){
if(a[j][i] == 'C')
tmp++;
}
ans += tmp*(tmp-1)/2;
}
printf("%d\n",ans);
}
return 0;
}

Far Relative’s Problem

题目链接：

http://www.codeforces.com/problemset/problem/629/B

解题思路：

Consider that we have boyi males
in the i'th day of the year and girli females
in the i'th day of the year. These arrays can be filled easily when you are reading the input (See the code). Then for the i'th day
of the year, we could have 2 * min(boyi , girli)
people which could come to the party. The answer would be maximum of this value between all days i (1 ≤ i ≤ 366)

Time Complexity: O(366*n)

AC代码：

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

struct node{
char sex[10];
int a,b;
}no[5005];

int main(){
int n;
while(~scanf("%d",&n)){
for(int i = 0; i < n; i++)
scanf("%s%d%d",no[i].sex,&no[i].a,&no[i].b);
int ans = 0,M,F;
for(int i = 1; i <= 366; i++){
M = 0;F = 0;
for(int j = 0; j < n; j++){
if(no[j].a <= i && no[j].b >= i){
if(no[j].sex[0] == 'M')
M++;
else
F++;
}
}
ans = max(ans,min(M,F));
}
printf("%d\n",ans*2);
}
return 0;
}

Famil Door and Brackets

题目链接：

http://www.codeforces.com/problemset/problem/629/C

解题思路：

This problem can be solved with dynamic programming:

1. Calculate dpi, j :
How many sequences of brackets of length i has balance j and
intermediate balance never goes below zero (They form a prefix of a valid sequence of brackets).

2. For the given sequence of length n calculate
the resulting balance a and the minimum balance b.

3. Try the length of the sequence added at the beginning c and
its balance d. If  - b ≤ d then
add dpc, d * dm - n - c, d + a to
the answer.

Time complexity: O((n - m)^2)

AC代码：

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

typedef long long ll;
const int MOD = 1e9+7;
ll dp[5005][5005];

int main(){
ll n,m;
while(~scanf("%lld%lld",&n,&m)){
string str;
cin>>str;
dp[0][0] = 1;
for(int i = 1; i <= n-m; i++){
dp[i][0] = dp[i-1][1];
for(int j = 1; j <= i; j++){
dp[i][j] = dp[i-1][j+1]+dp[i-1][j-1];
dp[i][j] %= MOD;
}
}
int num = 0,minnum = 0;
for(int i = 0; i < m; i++){
if(str[i] == '(')
num++;
else
num--;
minnum = min(minnum,num);
}
ll ans = 0;
for(int i = 0; i <= n-m; i++){
for(int j = 0; j <= i; j++){
if(j >= -minnum){
if(j+num <= n-m && j+num >= 0){
ans += dp[i][j] * dp[n - m - i][j + num] % MOD;
ans %= MOD;
}
}
}
}
printf("%lld\n",ans);
}
return 0;
}

Babaei and Birthday Cake

题目链接：

http://www.codeforces.com/problemset/problem/629/D

解题思路：

First of all, we calculate the volume of each cake: vi = π × hi × ri2.

Now consider the sequence v1, v2, v3,
..., vn :
The answer to the problem is the maximum sum of elements between all increasing sub-sequences of this sequence. How do we solve this? First to get rid of the decimals we can define a new sequence a1, a2, a3,
..., an such
that 


We consider dpi as
the maximum sum between all the sequences which end with ai and

dpi = 


The answer to the problem is: π × maxi = 1ndp[i]

Now how do we calculate 

 ?
We use a max-segment tree which does these two operations: 1. Change the i't member
to v. 2. Find
the maximum value in the interval 1 to i.

Now we use this segment tree for the array dp and find the answer.

Consider that a1, a2, a3,
..., an is
sorted. We define bi as
the position of ai.
Now to fill dpi we
find the maximum in the interval [1, bi) in
segment and we call it x and we set the bi th
index of the segment as ai + x.
The answer to the problem would the maximum in the segment in the interval [1,n]

Time complexity: O(nlgn)

AC代码：

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

typedef long long ll;
const int maxn = 100005;
const double pi = acos(-1.0);

struct node{
ll r, h, v;
}no[maxn];
struct node2{
int l,r;
ll maxv;
}tree[maxn<<2];
int n,m;
ll d[maxn],b[maxn];

void pushup(int id) {
tree[id].maxv = max(tree[id<<1].maxv,tree[id<<1|1].maxv);
}

void build(int id,int l,int r){
tree[id].l = l;
tree[id].r = r;
tree[id].maxv = 0;
if(l == r)
return ;
int m = (l + r) >> 1;
build(id<<1,l,m);
build(id<<1|1,m+1,r);
}

void update(int id,int l,int r,int val){
if(tree[id].l >= l && tree[id].r <= r){
if(tree[id].maxv < d[val]){
tree[id].maxv = d[val];
}
return ;
}
int mid = (tree[id].l+tree[id].r)>>1;
if(l <= mid)
update(id<<1,l,r,val);
if(mid < r)
update(id<<1|1,l,r,val);
pushup(id);
}

ll query(int id,int l,int r) {
if(tree[id].l >= l && tree[id].r <= r)
return tree[id].maxv;
ll ans = 0;
int mid = (tree[id].l+tree[id].r)>>1;
if(l <= mid)
ans = max(ans, query(id<<1,l,r));
if(mid< r)
ans = max(ans, query(id<<1|1,l,r));
pushup(id);
return ans;
}

int main(){
while(~scanf("%d",&n)){
for(int i = 1; i <= n; i++){
scanf("%lld%lld",&no[i].r,&no[i].h);
no[i].v = no[i].r * no[i].r * no[i].h;
b[i] = no[i].v;
}
sort(b+1, b+1+n);
int m = unique(b+1, b+1+n) - b - 1;

ll cur = 0,v;
build(1,1,m);
for(int i = 1; i <= n; i++) {
int L = lower_bound(b+1, b+1+m, no[i].v) - b;
if(L > 1) v = query(1,1,L-1);
else v = 0;
d[i] = (v + no[i].v);
update(1,L,L,i);
cur = max(cur, d[i]);
}
printf("%.9f\n",pi*cur);
}
return 0;
}