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首页 > 编程语言> > 第 45 届国际大学生程序设计竞赛(ICPC)亚洲区域赛(南京)Evil Coordinate (全排列)

第 45 届国际大学生程序设计竞赛(ICPC)亚洲区域赛(南京)Evil Coordinate (全排列)

作者:互联网

传送门

题意

给出一个字符串,炸弹的坐标;
从0,0点出发,按照字符串所给出的方向走,如果中途踩到炸弹,则需要重新排列字符串使路径中没有炸弹,输出排列后的路径字符串,否则输出“Impossible”,

思路1

这个是当时的思路,但是因为码力比较弱,没写对;

Code

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <string>
#include <algorithm>
#include <queue>
#include <utility>
#include <stack>
#include <map>
#include <vector>
#include <set>
#include <iomanip>
#define hz020 return
#define mes memset
#define mec memcpy

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int>pii;

const int N = 1000010;
const int null = 0x3f3f3f3f;
const ll mod = 1000000007;

int T;
int x,y; 

int main()
{
	scanf("%d",&T);
	while(T --)
	{
		cin >> x >> y;
		string s;
		cin >> s;
		int u = 0,d = 0,l = 0,r = 0,idx = 0,idy = 0; 
		for(int i = 0;i < s.size();i ++)
		{
			if(s[i] == 'U') u ++,idy ++;
			else if(s[i] == 'D') d ++,idy --;
			else if(s[i] == 'R') r ++,idx ++;
			else if(s[i] == 'L') l ++,idx --;
		//	cout << idx << " " << idy << endl;
		}
	//	cout << idx << endl << idy << endl;
		if((x == idx && y == idy) || (x == 0 && y == 0)) puts("Impossible");
		else if(idx == x && idy != y)//终点和炸弹都在一条竖线上
		{
			if(idx == 0 && l == 0 && r == 0)//y轴上并且不可以左右走 
			{
				if(idy >= y && y >= 0 && idy >= 0) puts("Impossible");//y正半轴,炸弹在终点下方
				else if(idy <= y && y <= 0 && idy <= 0) puts("Impossible");//y负半轴,炸弹在终点上方 
				else if(y < 0)//炸弹在负半轴 
				{
					for(int i = 1;i <= u;i ++) cout << "U";
					for(int i = 1;i <= d;i ++) cout << "D";
					puts("");
					
				}
				else if(y > 0)//炸弹在正半轴 
				{
					for(int i = 1;i <= d;i ++) cout << "D";
					for(int i = 1;i <= u;i ++) cout << "U";
					puts(""); 
				}
			}
			else if(idx == 0)//终点和炸弹都在y轴上,但是左右步数不为0 
			{
				for(int i = 1;i <= l;i ++) cout << "L";
				for(int i = 1;i <= u;i ++) cout << "U";
				for(int i = 1;i <= d;i ++) cout << "D";
				for(int i = 1;i <= r;i ++) cout << "R";
				puts("");	
			}
			else//终点和炸弹在一条竖线上且不在坐标轴上 
			{
				for(int i = 1;i <= d;i ++) cout << "D";
				for(int i = 1;i <= u;i ++) cout << "U";
				for(int i = 1;i <= l;i ++) cout << "L";
				for(int i = 1;i <= r;i ++) cout << "R";
				
				puts("");
			} 
		}
		else if(idy == y && idx != x) //终点和炸弹在一条直线上
		{
			if(idy == 0 && u == 0 && d == 0)//终点和炸弹在x轴上,且不可以上下移动 
			{
				if(idx >= x && x >= 0 && idx >= 0) puts("Impossible");//x正半轴,炸弹在终点左侧 
				else if(idx <= x && idx <= 0 && x <= 0) puts("Impossible");//x负半轴,炸弹在终点右侧 
				else if(x > 0)//炸弹在正半轴 
				{
					for(int i = 1;i <= l;i ++) cout << "L";
					for(int i = 1;i <= r;i ++) cout << "R";
					puts("");
				} 
				else if(x < 0) //炸弹在负半轴 
				{
					for(int i = 1;i <= r;i ++) cout << "R";
					for(int i = 1;i <= l;i ++) cout << "L";
					puts("");
				} 
			}
			else if(idy == 0) //炸弹和终点都在x轴但是可以上下走 
			{
				for(int i = 1;i <= u;i ++) cout << "U";
				for(int i = 1;i <= l;i ++) cout << "L";
				for(int i = 1;i <= r;i ++) cout << "R";
				for(int i = 1;i <= d;i ++) cout << "D";
				puts("");
			}
			else//终点和炸弹在一条横线上且不在坐标轴上 
			{
				for(int i = 1;i <= l;i ++) cout << "L";
				for(int i = 1;i <= r;i ++) cout << "R";
				for(int i = 1;i <= d;i ++) cout << "D";
				for(int i = 1;i <= u;i ++) cout << "U";
				puts("");
			}
			
		} 
		else
		{
			if(y == 0) // 在剩下的横轴 
			{
				for(int i = 1;i <= d;i ++) cout << "D";
				for(int i = 1;i <= u;i ++) cout << "U";
				for(int i = 1;i <= l;i ++) cout << "L";
				for(int i = 1;i <= r;i ++) cout << "R";
				
				puts("");
			}
			else // 在剩下的竖轴 
			{
				for(int i = 1;i <= l;i ++) cout << "L";
				for(int i = 1;i <= r;i ++) cout << "R";
				for(int i = 1;i <= d;i ++) cout << "D";
				for(int i = 1;i <= u;i ++) cout << "U";
				puts("");
			}
		}
	} 
	
	
	hz020 0;
}  

思路2

因为炸弹只有一个点,我们将上下左右规划到一个点,将四个方向进行全排列,在每一次排列中判断路径上是否有地雷;

Code

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <string>
#include <algorithm>
#include <queue>
#include <utility>
#include <stack>
#include <map>
#include <vector>
#include <set>
#include <iomanip>
#define hz020 return
#define mes memset
#define mec memcpy

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int>pii;

const int N = 1000010;
const int null = 0x3f3f3f3f;
const ll mod = 1000000007;

int T;
int x,y; 
int cnt[210];//记录次数
int d[2][4] = {{0,1,0,-1},{1,0,-1,0}};

string str = "URDL";

int main()
{
	scanf("%d",&T);
	while(T --)
	{
		mes(cnt,0,sizeof cnt);
		cin >> x >> y;
		string s;
		cin >> s;
		for(auto i:s) cnt[i] ++;
		int idy = cnt[str[0]] - cnt[str[2]],idx = cnt[str[1]] - cnt[str[3]];
		if((x == idx && y == idy) || (x == 0 && y == 0))
		{
			puts("Impossible");
			continue;
		}
		int p[4] = {0,1,2,3};//代表方向
		do
		{
			idx = 0,idy = 0;
			string ss;
			for(int i = 0;i < 4;i ++)
			{
				for(int j = 0;j < cnt[str[p[i]]];j ++)
				{
					ss += str[p[i]];
					idx += d[0][p[i]];
					idy += d[1][p[i]];
					if(idx == x && idy == y) goto pjq;
				} 
			}
			
			cout << ss << endl;
			goto pjq0200;
			
			pjq:; 
		}while(next_permutation(p,p + 4));
		
		cout << "Impossible" << endl;
		
		pjq0200:;
		
	} 
	
	
	hz020 0;
}  

标签:idx,idy,int,45,Evil,ICPC,++,&&,include
来源: https://blog.csdn.net/m0_52348473/article/details/116698030