标签:bool find 要求 getch amp inline using oid main
题目大意:
给定\(n\)个蚂蚁和\(n\)颗苹果树的坐标,要求每个蚂蚁爬到一颗苹果树旁,使得每个蚂蚁路线不相交且路线总长度最小,求每个蚂蚁爬到哪个苹果树旁?
首先假设有两只蚂蚁路径相交,那么这两个蚂蚁交换目标一定使得总路线缩短且不相交,所以总长度最短时所有蚂蚁路线一定不相交
怎么让总路线最短呢?二分图最小权匹配
其实只要把边权全部取反然后跑最大权匹配就好了
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
namespace red{
#define int long long
#define eps (1e-6)
inline int read()
{
int x=0;char ch,f=1;
for(ch=getchar();(ch<'0'||ch>'9')&&ch!='-';ch=getchar());
if(ch=='-') f=0,ch=getchar();
while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return f?x:-x;
}
const int N=210;
int n;
struct node
{
int x,y;
}ant[N],tr[N];
double jx[N][N];
inline int sqr(int x){return x*x;}
inline double dis(int i,int j)
{
return sqrt(sqr(ant[i].x-tr[j].x)+sqr(ant[i].y-tr[j].y));
}
double exl[N],exr[N],slack[N];
bool visl[N],visr[N];
int f[N],g[N];
inline bool find(int x)
{
visl[x]=1;
for(int y=1;y<=n;++y)
{
if(visr[y]) continue;
double tmp=exl[x]+exr[y]-jx[x][y];
if(fabs(tmp)<eps)
{
visr[y]=1;
if(!f[y]||find(f[y]))
{
f[y]=x;
g[x]=y;
return 1;
}
}
else slack[y]=min(tmp,slack[y]);
}
return 0;
}
inline void km()
{
for(int i=1;i<=n;++i)
{
exl[i]=jx[i][1];
for(int j=2;j<=n;++j)
{
exl[i]=max(exl[i],jx[i][j]);
}
}
for(int i=1;i<=n;++i)
{
for(int j=1;j<=n;++j)
{
visl[j]=visr[j]=0;
slack[j]=1e9+7;
}
if(find(i)) continue;
while("haku")
{
double tmp=1e9+7;
int t;
for(int j=1;j<=n;++j)
if(!visr[j]) tmp=min(tmp,slack[j]);
for(int j=1;j<=n;++j)
{
if(visl[j]) exl[j]-=tmp;
if(visr[j]) exr[j]+=tmp;
else
{
slack[j]-=tmp;
if(fabs(slack[j])<eps) t=j;
}
}
if(!f[t]) break;
visr[t]=1,visl[f[t]]=1;
t=f[t];
for(int j=1;j<=n;++j)
slack[j]=min(slack[j],exl[t]+exr[j]-jx[t][j]);
}
memset(visl,0,sizeof(visl));
memset(visr,0,sizeof(visr));
find(i);
}
for(int i=1;i<=n;++i) printf("%lld\n",g[i]);
}
inline void main()
{
while(~scanf("%lld",&n))
{
memset(f,0,sizeof(f));
memset(g,0,sizeof(g));
memset(exl,0,sizeof(exl));
memset(exr,0,sizeof(exr));
for(int i=1;i<=n;++i)
{
ant[i].x=read(),ant[i].y=read();
}
for(int i=1;i<=n;++i)
{
tr[i].x=read(),tr[i].y=read();
}
for(int i=1;i<=n;++i)
{
for(int j=1;j<=n;++j)
{
jx[i][j]=-dis(i,j);
}
}
km();
}
}
}
signed main()
{
red::main();
return 0;
}
标签:bool find 要求 getch amp inline using oid main
原文地址:https://www.cnblogs.com/knife-rose/p/12093286.html