题目链接
https://vjudge.net/problem/Gym-102465D
题意
n*m矩阵,k个节点,选择一行从左到右走,当遇到同一列有节点时,要走过去再回来。求最短路径
思路
每一列上的x个节点可以看成长为len的线段,当选择的行穿过它时,这几个节点的花费时len*2,否则的话,需要额外再加上选择的行到线段端点距离的二倍。三分选择行,取最小值即可。
注意本题需要long long
代码
#include<cstdio>
#include<iostream>
#include<iomanip>
#include<map>
#include<string>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define endl "\n"
#define int long long
//#define double long double
using namespace std;
typedef long long ll;
const int maxn=100500;
const int inf=0x3f3f3f3f3f3f3f3f;
int n,m,k;
int x[maxn],y[maxn],high[maxn],low[maxn];
bool bl[maxn];
int fun(int i){
int rec=n-1;
for(int j=0;j<=n;j++){
if(!bl[j])
continue;
rec+=2*(high[j]-low[j]);
int d1=i-low[j],d2=high[j]-i;
if(d1<0)
rec-=2*d1;
else if(d2<0)
rec-=2*d2;
}
return rec;
}
signed main(){
IOS
cin>>n>>m>>k;
memset(high,0,sizeof high);
memset(low,0x3f,sizeof low);
for(int i=1;i<=k;i++){
cin>>x[i]>>y[i];
bl[x[i]]=1;
high[x[i]]=max(high[x[i]],y[i]);
low[x[i]]=min(low[x[i]],y[i]);
}
int ans=inf;
int l=-1,r=m+1,lm,rm;
for(int i=1;i<=100;i++){
lm=l+(r-l)/3;
rm=r-(r-l)/3;
int a1=fun(lm),a2=fun(rm);
ans=min(ans,min(a1,a2));
if(a1<a2)
r=rm;
else
l=lm;
}
for(int i=l;i<=r;i++)
ans=min(ans,fun(i));
cout<<ans<<endl;
return 0;
}