CF 82 D.Two out of Three

前言

全网唯一不同题解

设 (f[i][j]) 表示第 (i) 次选取留下来的数是 (k) 的最小花费

枚举前面的留下来的点 (k) 当前能留下的点只有 ((2*i),(2*i+1),k) 中的一个，时间复杂度 (O(n^2))

选取次数是 (n/2) 向上取整。因为最后什么点也不留下，(a[n+1]=0)，所以答案可以为 (f[m][n+1](m=n/2向上取整))

另外我没有写记录路径，记录一下也挺简单

Code

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;
inline int read() {
	int 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<<3)+(x<<1)+(ch^48); ch=getchar(); }
	return x * f;
}
const int N = 1007;
int n;
int a[N];
int f[N][N];
inline int Const(int x,int y) {
	return max(a[x], a[y]);
}
int main()
{
	n = read();
	for(int i=1;i<=n;++i)
		a[i] = read();
	memset(f, 0x3f, sizeof(f));
	f[1][1] = max(a[2],a[3]); f[1][2] = max(a[1],a[3]); f[1][3] = max(a[1],a[2]);
	int m = n&1 ? n/2+1 : n/2;
	for(int i=2;i<=m;++i) {
		int x = 2*i, y = 2*i+1;	// i次选取的最后两个数 
		for(int k=1;k<x;++k) {	//枚举上一次留下的数 
			f[i][x] = min(f[i][x], f[i-1][k]+Const(y,k));
			f[i][y] = min(f[i][y], f[i-1][k]+Const(x,k));
			f[i][k] = min(f[i][k], f[i-1][k]+Const(x,y));
		}
	}
	int ans = f[m][n+1];
	printf("%d
",ans);
	return 0;
}

update

加上路径记录版

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;
inline int read() {
	int 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<<3)+(x<<1)+(ch^48); ch=getchar(); }
	return x * f;
}
const int N = 1007;
int n;
int a[N];
int f[N][N];
struct operation {
	int x,y,pre;
}p[N][N];
inline int Const(int x,int y) {
	return max(a[x], a[y]);
}
void output(int now,int pos) {
	if(now == 0) return ;
	output(now-1,p[now][pos].pre);
	if(a[p[now][pos].x])	
		printf("%d ",p[now][pos].x);
	if(a[p[now][pos].y])
		printf("%d",p[now][pos].y);
	puts("");
}
int main()
{
	n = read();
	for(int i=1;i<=n;++i)
		a[i] = read();
	memset(f, 0x3f, sizeof(f));
	f[1][1] = max(a[2],a[3]); f[1][2] = max(a[1],a[3]); f[1][3] = max(a[1],a[2]);
	p[1][1] = (operation)<%2,3,0%>;
	p[1][2] = (operation)<%1,3,0%>;
	p[1][3] = (operation)<%1,2,0%>;
	int m = n&1 ? n/2+1 : n/2;
	for(int i=2;i<=m;++i) {
		int x = 2*i, y = 2*i+1;	// i次选取的最后两个数 
		for(int k=1;k<x;++k) {	//枚举上一次留下的数 
			if(f[i-1][k]+Const(y,k) < f[i][x]) {
				f[i][x] = f[i-1][k]+Const(y,k);
				p[i][x] = (operation)<%k,y,k%>;
			}
			if(f[i-1][k]+Const(x,k) < f[i][y]) {
				f[i][y] = f[i-1][k]+Const(x,k);
				p[i][y] = (operation)<%k,x,k%>;
			}
			if(f[i-1][k]+Const(x,y) < f[i][k]) {
				f[i][k] = f[i-1][k]+Const(x,y);
				p[i][k] = (operation)<%x,y,k%>;
			}
			// f[i][x] = min(f[i][x], f[i-1][k]+Const(y,k));
			// f[i][y] = min(f[i][y], f[i-1][k]+Const(x,k));
			// f[i][k] = min(f[i][k], f[i-1][k]+Const(x,y));
		}
	}
	int ans = f[m][n+1];
	printf("%d
",ans);
	output(m,n+1);
	return 0;
}