【LGR-225-Div.1.5】洛谷蓝桥杯模拟赛 G P12596 测速仪惊吓

思路讲解

赛时稍微有点暴力，其实这个是增长的，递增的（不要从代数上想，从几何的角度想就容易很多）。

那么二分或者是双指针都行，那我们就写一个双指针吧，好久没写了。

// 双指针做法
FOR(i,1,N-1){
	FOR(j,bp,N){
		ll curArea=Sum[j-1]-Sum[i-1];
		ll ldiff=Sum[N]-curArea*2;  // ldiff不能加abs，只要ldiff变为负数了，继续加j一定会变大
		if(ldiff>=0){
			ans=min(ans,abs(ldiff));
		}else{
			ans=min(ans,abs(ldiff));
			bp=j;
			break;
		}
	}
}

核心做法如上，最重要的就是ldiff不能加abs，只要 ldiff 变为负数了，继续加 j 一定会变大（绝对值）。

AC代码

https://www.luogu.com.cn/record/218460139

源代码

// Problem: T610809 测速仪惊吓
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/T610809?contestId=245376
// Memory Limit: 1 MB
// Time Limit: 2000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
    do { \
        std::cerr << #variable << ":"; \
        for (const auto& elem : variable) { \
            std::cerr << elem << " "; \
        } \
        std::cerr << "\n"; \
    } while (0)
#define uniVec(var) \
    do { \
        sort(var.begin(),var.end());\
        var.resize(unique(var.begin(),var.end())-var.begin());\
    } while (0)
#define debugN(var,N) \
	do{ \
		std::cerr<<#var<<":"; \
		FOR(i,1,N){ \
			std::cerr<<var[i]<<" "; \
		} \
		std::cerr<<"\n"; \
	}while(0)
#define debugMap(variable) \
    do { \
        std::cerr << #variable << ":\n"; \
        for (const auto& pair : variable) { \
            std::cerr << "  " << pair.first << " => " << pair.second << "\n"; \
        } \
        std::cerr << std::endl; \
    } while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define vecMxx(vec,x) (vec.end()-lower_bound(all(vec),x))	// vec中大于x的数有几个

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr3;
typedef array<ll,2> arr2;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e5)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,Q,X,K,T,A[MAXN];
arr2 xy[MAXN];
inline ll Cross(ll a,ll b,ll c,ll d){	// 记得多边形面积要除以2
	return a*d-b*c;
}
/*

*/
inline ll read()
{
	ll x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9')
		{
			if(ch=='-')
				f=-1;
			ch=getchar();
		}
	while(ch>='0' && ch<='9')
		x=x*10+ch-'0',ch=getchar();
	return x*f;
}
ll Sum[MAXN];
inline void solve(){
	N=read();
	FOR(i,1,N){
		ll x=read(),y=read();
		xy[i][0]=x;xy[i][1]=y;
	}
	ll px=read(),py=read();
	// Sum的逻辑变为三角形面积的前缀和了（以P和边的顶点组成的三角形）严格来说是其面积的两倍
	FOR(i,1,N){
		ll x1=xy[i][0],y1=xy[i][1],x2=xy[(i+1)<=N?i+1:1][0],y2=xy[(i+1)<=N?i+1:1][1];
		Sum[i]=Sum[max(i-1,0)]+abs(Cross(x1-px,y1-py,x2-px,y2-py));
	}
	// 我们面积就不除以2了
	ll ans=INF;
	ll bp=2;
	// 双指针做法
	FOR(i,1,N-1){
		FOR(j,bp,N){
			ll curArea=Sum[j-1]-Sum[i-1];
			ll ldiff=Sum[N]-curArea*2;
			if(ldiff>=0){
				ans=min(ans,abs(ldiff));
			}else{
				ans=min(ans,abs(ldiff));
				bp=j;
				break;
			}
		}
	}
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	// cin>>T;
	// while(T--){
		// solve();
	// }
	solve();
	return 0;
}
/*
AC
https://www.luogu.com.cn/record/218460139
*/

心路历程（WA，TLE，MLE……）

https://www.luogu.com.cn/record/218309162

赛时TLE，也骗到了80分，可以了可以了。

源代码

// Problem: T610809 测速仪惊吓
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/T610809?contestId=245376
// Memory Limit: 1 MB
// Time Limit: 2000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
    do { \
        std::cerr << #variable << ":"; \
        for (const auto& elem : variable) { \
            std::cerr << elem << " "; \
        } \
        std::cerr << "\n"; \
    } while (0)
#define uniVec(var) \
    do { \
        sort(var.begin(),var.end());\
        var.resize(unique(var.begin(),var.end())-var.begin());\
    } while (0)
#define debugN(var,N) \
	do{ \
		std::cerr<<#var<<":"; \
		FOR(i,1,N){ \
			std::cerr<<var[i]<<" "; \
		} \
		std::cerr<<"\n"; \
	}while(0)
#define debugMap(variable) \
    do { \
        std::cerr << #variable << ":\n"; \
        for (const auto& pair : variable) { \
            std::cerr << "  " << pair.first << " => " << pair.second << "\n"; \
        } \
        std::cerr << std::endl; \
    } while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define vecMxx(vec,x) (vec.end()-lower_bound(all(vec),x))	// vec中大于x的数有几个

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr3;
typedef array<ll,2> arr2;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e5)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,Q,X,K,T,A[MAXN];
arr2 xy[MAXN];
inline ll Cross(ll a,ll b,ll c,ll d){	// 记得多边形面积要除以2
	return a*d-b*c;
}
/*

*/
inline ll read()
{
	ll x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9')
		{
			if(ch=='-')
				f=-1;
			ch=getchar();
		}
	while(ch>='0' && ch<='9')
		x=x*10+ch-'0',ch=getchar();
	return x*f;
}
ll Sum[MAXN];
inline void solve(){
	N=read();
	FOR(i,1,N){
		ll x=read(),y=read();
		xy[i][0]=x;xy[i][1]=y;
	}
	ll px=read(),py=read();
	FOR(i,1,N-1){
		Sum[i]=Sum[i-1]+Cross(xy[i][0],xy[i][1],xy[i+1][0],xy[i+1][1]);
	}
	// 我们面积就不除以2了
	ll diff=INF;
	FOR(i,1,N){
		FOR(j,i+1,N){
			ll Crosspi=Cross(xy[i][0],xy[i][1],px,py);
			ll Crosspj=Cross(xy[j][0],xy[j][1],px,py);
			ll areaD1=Sum[N-1]-Sum[j-1]+Sum[i-1]+Cross(xy[N][0],xy[N][1],xy[1][0],xy[1][1])
			+Cross(px,py,xy[j][0],xy[j][1])+Cross(xy[i][0],xy[i][1],px,py);
			ll areaD2=Sum[j-1]-Sum[i-1]+Cross(px,py,xy[i][0],xy[i][1])+
			Cross(xy[j][0],xy[j][1],px,py);
			areaD1=abs(areaD1);areaD2=abs(areaD2);
			diff=min(diff,abs(areaD1-areaD2));
		}
	}
	cout<<diff<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	// cin>>T;
	// while(T--){
		// solve();
	// }
	solve();
	return 0;
}
/*

*/