【LGR-225-Div.1.5】洛谷蓝桥杯模拟赛 G P12596 测速仪惊吓 · znzryb
## 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。
```cpp
// 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](https://www.luogu.com.cn/record/218460139)
::: 源代码
```cpp
// 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](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
```cpp
// 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;
}
/*
*/
```
:::