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CodeForces 1296 C Yet Another Walking Robot

由 别来无恙 提交于 2020-02-06 03:08:56

在这里插入图片描述
在这里插入图片描述

题意：

有四个走的方向，问保证不改变初始位置和终点位置的情况下怎么删除最少的字符串才符合。
用一个map来记录走过的位置，只要出现回到当前位置我们就把这一段的给删掉。
两种写法。

Code1:Code1:

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}
 
ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}
 
ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
    if (a >= mod)
        a = a % mod + mod;
    ll ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ans * a;
            if (ans >= mod)
                ans = ans % mod + mod;
        }
        a *= a;
        if (a >= mod)
            a = a % mod + mod;
        b >>= 1;
    }
    return ans;
}
 
// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}
 
///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}
 
///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}
 
///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}
const int N = 2e5 + 5;
int n;
int l = -1, r = 1, u = 1, d = -1, sum1[N], sum2[N];
map<pair<int, int>, int> mp;
int ans, ans1, ans2;
 
int main()
{
    int t;
    sd(t);
    sum1[0] = 0;
    sum2[0] = 0;
    while (t--)
    {
        sd(n);
        getchar();
        mp.clear();
        ans = n + 1;
        mp[{0, 0}] = 0;
        rep(i, 1, n)
        {
            char c = getchar();
            sum1[i] = sum1[i - 1];
            sum2[i] = sum2[i - 1];
            if (c == 'L')
                sum1[i] += l;
            else if (c == 'R')
                sum1[i] += r;
            else if (c == 'U')
                sum2[i] += u;
            else
                sum2[i] += d;
            if (mp.count({sum1[i], sum2[i]}))
            {
                if (i - mp[{sum1[i], sum2[i]}] < ans)
                {
                    ans = i - mp[{sum1[i], sum2[i]}];
                    ans1 = mp[{sum1[i], sum2[i]}] + 1, ans2 = i;
                }
            }
            mp[{sum1[i], sum2[i]}] = i;
        }
        if (ans == n + 1)
            puts("-1");
        else
            pdd(ans1, ans2);
    }
    return 0;
}

Code2:Code2:

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
	int ret = 0, sgn = 1;
	char ch = getchar();
	while (ch < '0' || ch > '9')
	{
		if (ch == '-')
			sgn = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9')
	{
		ret = ret * 10 + ch - '0';
		ch = getchar();
	}
	return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
	if (a > 9)
		Out(a / 10);
	putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
	return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
	return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
	if (a >= mod)
		a = a % mod + mod;
	ll ans = 1;
	while (b)
	{
		if (b & 1)
		{
			ans = ans * a;
			if (ans >= mod)
				ans = ans % mod + mod;
		}
		a *= a;
		if (a >= mod)
			a = a % mod + mod;
		b >>= 1;
	}
	return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
	return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
	if (b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}
	int g = exgcd(b, a % b, x, y);
	int t = x;
	x = y;
	y = t - a / b * y;
	return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
	int d, x, y;
	d = exgcd(a, p, x, y);
	if (d == 1)
		return (x % p + p) % p;
	else
		return -1;
}

///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
	int M = 1, y, x = 0;
	for (int i = 0; i < n; ++i) //算出它们累乘的结果
		M *= a[i];
	for (int i = 0; i < n; ++i)
	{
		int w = M / a[i];
		int tx = 0;
		int t = exgcd(w, a[i], tx, y); //计算逆元
		x = (x + w * (b[i] / t) * x) % M;
	}
	return (x + M) % M;
}
const int N = 2e5 + 5;
int n;
ll l = -1, r = 1, u = inf, d = -inf, sum[N];
map<ll, int> mp;
int ans, ans1, ans2;
int main()
{
	int t;
	sd(t);
	sum[0] = 0;
	while (t--)
	{
		sd(n);
		getchar();
		mp.clear();
		ans = n + 1;
		mp[0] = 0;
		rep(i, 1, n)
		{
			char c = getchar();
			sum[i] = sum[i - 1];
			if (c == 'L')
				sum[i] += l;
			else if (c == 'R')
				sum[i] += r;
			else if (c == 'U')
				sum[i] += u;
			else
				sum[i] += d;
			if (mp.count(sum[i]))
			{
				if (i - mp[sum[i]] < ans)
				{
					ans = i - mp[sum[i]];
					ans1 = mp[sum[i]] + 1, ans2 = i;
				}
			}
			mp[sum[i]] = i;
		}
		if (ans == n + 1)
			puts("-1");
		else
			cout << ans1 << " " << ans2 << endl;
	}
	return 0;
}
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