编程语言
首页 > 编程语言> > #KM算法#UVA11383 Golden Tiger Claw

#KM算法#UVA11383 Golden Tiger Claw

作者:互联网

题目

给定 \(n*n\) 的矩阵,现在给每行安排一个权值 \(x_i\),给每列安排一个权值 \(y_j\),

使得 \(x_i+y_j\geq a_{i,j}\),并且使 \(\sum_{i=1}^nx_i+y_i\) 最小。


分析

学过KM算法的话,就应该知道可以将 \(x_i\) 和 \(y_i\) 当成顶标,并且当 \(x_i+y_j=a_{i,j}\) 时取得最小值,

那么就转换成二分图最大权匹配,直接跑KM算法将匹配的边权和加起来就是答案。


代码

#include <cstdio>
#include <cctype>
#include <cmath>
#include <queue>
using namespace std;
const int N=511;
bool vx[N],vy[N]; queue<int>q;
int slack[N],lx[N],ly[N],G[N][N];
int px[N],py[N],link[N],n,ans;
int iut(){
	int ans=0,f=1; char c=getchar();
	while (!isdigit(c)) f=(c=='-')?-f:f,c=getchar();
	while (isdigit(c)) ans=ans*10+c-48,c=getchar();
	return ans*f;
}
void print(int ans){
	if (ans>9) print(ans/10);
	putchar(ans%10+48);
}
int min(int a,int b){return a<b?a:b;}
int max(int a,int b){return a>b?a:b;}
void adjust(int y){
	for (int _y;y;y=_y){
		_y=px[link[y]];
		px[link[y]]=y;
		py[y]=link[y]; 
	}
}
void bfs(int st){
	for (int i=1;i<=n;++i) slack[i]=0x3f3f3f3f,vx[i]=vy[i]=0;
	while (!q.empty()) q.pop();
	q.push(st);
	while (1){
		while (!q.empty()){
			int x=q.front();
			vx[x]=1,q.pop();
			for (int y=1;y<=n;++y)
			if (!vy[y]&&slack[y]>lx[x]+ly[y]-G[x][y]){
				slack[y]=lx[x]+ly[y]-G[x][y],link[y]=x;
				if (!slack[y]){
					vy[y]=1;
					if (!py[y]) {adjust(y); return;}
					    else q.push(py[y]);
				}
			}
		}
		int mn=0x3f3f3f3f;
		for (int i=1;i<=n;++i)
		    if (!vy[i]) mn=min(mn,slack[i]);
		for (int i=1;i<=n;++i){
			if (vx[i]) lx[i]-=mn;
		    if (vy[i]) ly[i]+=mn;
		        else slack[i]-=mn;
		}
		for (int i=1;i<=n;++i)
		if (!vy[i]&&!slack[i]){
			vy[i]=1;
			if (!py[i]) {adjust(i); return;}
			    else q.push(py[i]);
		}
	}
}
void KM(){
	for (int i=1;i<=n;++i){
		link[i]=ly[i]=px[i]=py[i]=0,lx[i]=-0x3f3f3f3f;
		for (int j=1;j<=n;++j)
		    lx[i]=max(lx[i],G[i][j]);
	}
	for (int i=1;i<=n;++i) bfs(i);
}
int main(){
	while (scanf("%d",&n)==1){
		for (int i=1;i<=n;++i)
		for (int j=1;j<=n;++j) G[i][j]=iut();
		KM(),ans=0;
		for (int i=1;i<=n;++i) ans+=lx[i],ans+=ly[i];
		for (int i=1;i<=n;++i) print(lx[i]),putchar(i==n?10:32);
		for (int i=1;i<=n;++i) print(ly[i]),putchar(i==n?10:32);
		print(ans),putchar(10);
	}
	return 0;
}

标签:Golden,10,int,py,UVA11383,KM,link,ans,include
来源: https://www.cnblogs.com/Spare-No-Effort/p/15957069.html