本文详细解析了 CodeForces 平台上的四道题目：国王最短路径、货车装载、井字游戏和最小成本括号序列，提供了完整的代码实现和解题思路。
文章目录
A - Shortest path of the king
CodeForces - 3A
The king is left alone on the chessboard. In spite of this loneliness, he doesn’t lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, he wants to get from his current position s to square t in the least number of moves. Help him to do this.
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).
Input
The first line contains the chessboard coordinates of square s, the second line — of square t.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.
Output
In the first line print n — minimum number of the king’s moves. Then in n lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.
Examples
Input
a8
h1
Output
7
RD
RD
RD
RD
RD
RD
RD
感觉这道题还是挺水的，直接根据两个位置的坐标判断即可。
一开始while循环的条件||写成了&&，错了一次。
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char arr[1005];
int main(void)
{
char a, b, c, d;
int index = 0, cnt = 0;//存储的字符数、走的步数
scanf("%c%c", &a, &b);
getchar();
scanf("%c%c", &c, &d);
getchar();
while (a != c || b != d)
{
if (a < c)
{
arr[index++] = 'R';
a++;
}
else if (a > c)
{
arr[index++] = 'L';
a--;
}
if (b < d)
{
arr[index++] = 'U';
b++;
}
else if (b > d)
{
arr[index++] = 'D';
b--;
}
arr[index++] = '\n';
cnt++;
}
printf("%d\n", cnt);
for(int i = 0; i < index; i++)
printf("%c", arr[i]);
return 0;
}
B - Lorry
CodeForces - 3B
A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It’s known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body.
Input
The first line contains a pair of integer numbers n and v (1 ≤ n ≤ 10^5; 1 ≤ v ≤ 10^9), where n is the number of waterborne vehicles in the boat depot, and v is the truck body volume of the lorry in cubic metres. The following n lines contain the information about the waterborne vehicles, that is a pair of numbers ti, pi (1 ≤ ti ≤ 2; 1 ≤ pi ≤ 10^4), where ti is the vehicle type (1 – a kayak, 2 – a catamaran), and pi is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file.
Output
In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them.
Examples
Input
3 2
1 2
2 7
1 3
Output
7
2
先对大小为1和2的船分别进行排序，再取所有能取的1船，然后再依次判断，如果一个2船的容量更大的话就把1船换成2船，可以是一个2船换两个1船（没有空位置了且一个2船大于两个1船）、一个2船换一个1船（有一个空位置且一个2船大于一个1船）、一个2船换一个2船（有没有空位置均可且一个2船大于一个2船）。
感觉我不适合做这种题，错误不好找啊，错了无数次之后终于过了
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 1e5 + 5;
typedef struct
{
int pi;
int id;
}ship;
bool cmp(ship a, ship b)
{
return a.pi > b.pi;
}
ship arr1[N], arr2[N];
int ID[2 * N];
int main(void)
{
int n, v, t, p;
int now, num1, num2, ans;//目前占用的空间大小，目前使用的两种船的数量，最大的容量
int index1, index2;//给出的两种船的数量
index1 = index2 = 0;
scanf("%d%d", &n, &v);
for (int i = 0; i < n; i++)
{
scanf("%d%d", &t, &p);
if (t == 1)
{
arr1[index1].pi = p;
arr1[index1].id = i + 1;
index1++;
}
else
{
arr2[index2].pi = p;
arr2[index2].id = i + 1;
index2++;
}
}
sort(arr1, arr1 + index1, cmp);
sort(arr2, arr2 + index2, cmp);
now = num1 = num2 = ans = 0;
for (int i = 0; i < index1; i++)//先装1的
{
if (now == v)//装满了
break;
ans += arr1[i].pi;
now++;
ID[num1++] = arr1[i].id;
}
for (int i = 0; i < index2; i++)//再装2的
{
if (v - now < 2)//不能再装了
break;
ans += arr2[i].pi;
now += 2;
ID[num1 + num2] = arr2[i].id;
num2++;
}
int num = num1 + num2;
for (int i = num2; i < index2; i++)
{
if (v - now == 0 && num1 >= 2 && arr2[i].pi > arr1[num1 - 1].pi + arr1[num1 - 2].pi)//一个2换两个1
{
ans += arr2[i].pi - (arr1[num1 - 1].pi + arr1[num1 - 2].pi);
ID[num++] = arr2[i].id;
ID[num1 - 1] = ID[num1 - 2] = 0;
num1 -= 2;
num2++;
}
else if (v - now == 1 && num1 >= 1 && arr2[i].pi > arr1[num1 - 1].pi)// 一个2换一个1
{
ans += arr2[i].pi - arr1[num1 - 1].pi;
ID[num1 - 1] = arr2[i].id;
now++;
num1--;
num2++;
}
else if (num2 >= 1 && arr2[i].pi > arr2[num2 - 1].pi)//一个2换一个2
{
ans += arr2[i].pi - arr2[num2 - 1].pi;
ID[num2 - 1] = arr2[i].id;
}
else
break;
}
sort(ID, ID + num);
printf("%d\n", ans);
for (int i = 0; i < num; i++)
{
if (ID[i] != 0)
{
if (i != num - 1)
printf("%d ", ID[i]);
else
printf("%d\n", ID[i]);
}
}
return 0;
}
C - Tic-tac-toe
CodeForces - 3C
题解：https://blog.youkuaiyun.com/liuke19950717/article/details/51628792?utm_source=blogxgwz6
D - Least Cost Bracket Sequence
CodeForces - 3D
This is yet another problem on regular bracket sequences.
A bracket sequence is called regular, if by inserting “+” and “1” into it we get a correct mathematical expression. For example, sequences “(())()”, “()” and “(()(()))” are regular, while “)(”, “(()” and “(()))(” are not. You have a pattern of a bracket sequence that consists of characters “(”, “)” and “?”. You have to replace each character “?” with a bracket so, that you get a regular bracket sequence.
For each character “?” the cost of its replacement with “(” and “)” is given. Among all the possible variants your should choose the cheapest.
Input
The first line contains a non-empty pattern of even length, consisting of characters “(”, “)” and “?”. Its length doesn’t exceed 5·104. Then there follow m lines, where m is the number of characters “?” in the pattern. Each line contains two integer numbers ai and bi (1 ≤ ai, bi ≤ 106), where ai is the cost of replacing the i-th character “?” with an opening bracket, and bi — with a closing one.
Output
Print the cost of the optimal regular bracket sequence in the first line, and the required sequence in the second.
Print -1, if there is no answer. If the answer is not unique, print any of them.
Examples
Input
(??)
1 2
2 8
Output
4
()()
pair作优先队列的元素，先按第一个元素排序，如果相等就比较第二个元素。
#include <cstdio>
#include <iostream>
#include <queue>
#include <string>
using namespace std;
typedef long long ll;
priority_queue <pair<ll, int> > pque;
string s;
int main(void)
{
cin >> s;
if (s.size() % 2 == 1)
return puts("-1"), 0;
int cnt = 0;
ll ans = 0;
for (int i = 0; i < s.size(); i++)
{
if (s[i] == '(')
cnt++;
else if (s[i] == ')')
cnt--;
else//问号
{
int a, b;
scanf("%d%d", &a, &b);
cnt--; s[i] = ')'; ans += b;
pque.push(make_pair(b - a, i));//只存储问号
}
if (cnt < 0)
{
if (pque.size() == 0)//如果前面没有问号
break;
pair<int, int> x = pque.top(); pque.pop();
ans -= x.first; s[x.second] = '(';
cnt += 2;
}
}
if(cnt != 0)
return puts("-1"), 0;
cout << ans << endl;
cout << s << endl;
return 0;
}
题解：https://www.cnblogs.com/qscqesze/p/5266725.html
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