题解【洛谷P2070】刷墙

题面

将每一次移动的距离进行差分，前缀和判断移动的距离是否(geq 2)即可。

#include <bits/stdc++.h>
#define itn int
#define gI gi

using namespace std;

typedef long long ll;

inline int gi()
{
	int f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

inline ll gl()
{
	ll f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int maxn = 100003;

int n, ans, now, sum;
struct Node
{
	int x, cf;
} a[maxn * 2];

inline bool cmp(Node x, Node y) {return x.x < y.x;}

int main()
{
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	n = gi();
	for (int i = 1; i <= n; i+=1)
	{
		int x = gi(); 
		char s;
		scanf("%c", &s);
		if (s == 'L')
		{
			a[i * 2 - 1] = (Node){now, -1};
			a[i << 1] = (Node){now - x, 1};
			now -= x;
		}
		else 
		{
			a[i * 2 - 1] = (Node){now, 1};
			a[i << 1] = (Node){now + x, -1};
			now += x;
		}
	}
	sort(a + 1, a + 1 + (n << 1), cmp);
	now = a[1].cf;
	for (int i = 2; i <= n * 2; i+=1)
	{
		if (now >= 2) ans += a[i].x - a[i - 1].x;
		now += a[i].cf;
	}
	printf("%d
", ans);
	return 0;
}