CodeForces-1294B排序+pair使用

题目

There is a robot in a warehouse and n packages he wants to collect. The warehouse can be represented as a coordinate grid. Initially, the robot stays at the point (0,0). The i-th package is at the point (xi,yi). It is guaranteed that there are no two packages at the same point. It is also guaranteed that the point (0,0) doesn’t contain a package.
The robot is semi-broken and only can move up (‘U’) and right (‘R’). In other words, in one move the robot can go from the point (x,y) to the point (x+1,y) or to the point (x,y+1).
As we say above, the robot wants to collect all n packages (in arbitrary order). He wants to do it with the minimum possible number of moves. If there are several possible traversals, the robot wants to choose the lexicographically smallest path.

The string s of length n is lexicographically less than the string t of length n if there is some index 1≤j≤n that for all i from 1 to j−1 si=ti and sj<tj. It is the standard comparison of string, like in a dictionary. Most programming languages compare strings in this way.
Input
The first line of the input contains an integer t (1≤t≤100) — the number of test cases. Then test cases follow.
The first line of a test case contains one integer n (1≤n≤1000) — the number of packages.
The next n lines contain descriptions of packages. The i-th package is given as two integers xi and yi (0≤xi,yi≤1000) — the x-coordinate of the package and the y-coordinate of the package.
It is guaranteed that there are no two packages at the same point. It is also guaranteed that the point (0,0) doesn’t contain a package.
The sum of all values n over test cases in the test doesn’t exceed 1000.
Output
Print the answer for each test case.
If it is impossible to collect all n packages in some order starting from (0,0), print “NO” on the first line.
Otherwise, print “YES” in the first line. Then print the shortest path — a string consisting of characters ‘R’ and ‘U’. Among all such paths choose the lexicographically smallest path.
Note that in this problem “YES” and “NO” can be only uppercase words, i.e. “Yes”, “no” and “YeS” are not acceptable.

Input
3
5
1 3
1 2
3 3
5 5
4 3
2
1 0
0 1
1
4 3
Output
YES
RUUURRRRUU
NO
YES
RRRRUUU

题意与思路

机器人捡包，只能往右，往上走，问是否能够捡到所有包。
这里不能捡到的情况是：x大，y却小，即位于右下方的包。

按坐标排序，然后作差即为需要移动的步数

这里使用三个参数的sort()，学习一下cmp的用法。

//cmp：升序使用小于号，降序使用大于号

并且点对pair<int ,int >的知识复习一下。

AC代码

#include<bits/stdc++.h>
using namespace std;//排序规则：横坐标先排，纵坐标后排
bool cmp(pair <int,int>a,pair<int,int>b)
{if(a.first!=b.first)return a.first<b.first;return a.second<b.second;//先排横坐标，后排纵坐标
}
int main()
{int T;vector<pair<int,int> >v;cin>>T;while(T--){int n,x,y;//坐标 cin>>n;int flag=0;v.clear();//输入到vector中v.push_back(make_pair(0,0));for(int i=0;i<n;i++){    cin>>x>>y;v.push_back(make_pair(x,y));}sort(v.begin(),v.end(),cmp);//for(vector<pair<int,int> > ::iterator it=v.begin()+1;it!=v.end();it++)//迭代器遍历{//错误情况：位于右下方的点 if(it->first>(it-1)->first &&it->second<(it-1)->second){cout<<"NO"<<endl;flag=1;break;//跳出for循环  }           } if(flag) continue;//继续下一个例子cout<<"YES"<<endl;//先右后上，遍历/* for(vector<pair<int,int> >::iterator it=v.begin();it!=v.end();it++){cout<<it->first<<"  "<<it->second<<endl;}*/for(vector<pair<int,int> >::iterator it=v.begin()+1;it!=v.end();it++){if(it->first>(it-1)->first)//横坐标不同{int a=it->first-(it-1)->first;//作差，输出R个数while(a--)cout<<"R";}if(it->second>(it-1)->second)//纵坐标不同{int b=it->second-(it-1)->second;//输出U的个数while(b--)cout<<"U";}} cout<<endl;v.clear();}return 0;
}

测试结果

参考资料
https://vjudge.net/contest/359083#problem/C

希望对你有帮助。