[区间DP] Tinkoff Challenge - Elimination Round D题

Posted 2022-03-02 晁棠

NamomoCamp Daily 2

codeforces原题网址https://codeforces.com/problemset/problem/793/D

代码源oj网址http://oj.daimayuan.top/problem/437

题解：

两个地方的条件有点小小的不同，一个是开区间一个是闭区间，但是没什么影响，而且写法也可以规避出现环的情况。

题解视频

注意

记忆化搜索写法会更慢。

代码：

DP

// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm> 
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
//#define int long long

using namespace std;

const int N = 103, M = 2020;
int n, T, k, m;
int pi, p[N], nxt[M], to[M], cost[M];

inline void add_in(int u, int v, int c) 
	pi++; nxt[pi] = p[u]; p[u] = pi; to[pi] = v; cost[pi] = c;


void ready()

	IOS;
	cin >> n >> k >> m;
	//ffor(i, 1, n) ffor(j, 1, n) ma[i][j] = INF;
	ffor(i, 1, m) 
		int u, v, c;
		cin >> u >> v >> c;
		add_in(v, u, c);
		//ma[u][v] = min(ma[u][v], c);
	


struct DP 
	int L, R, Cnt, Flag;
;

int dp[N][N][N][3];
bool vis[N][N][N][3];

void work()

	mst(dp, inf);
	int ans = inf;
	queue<DP> q;
	ffor(i, 1, n) 
		q.push( i,i,1,0 );
		q.push( i,i,1,1 );
		dp[i][i][1][0] = dp[i][i][1][1] = 0;
		vis[i][i][1][0] = vis[i][i][1][1] = true;
	
	while (q.size()) 
		DP U = q.front(); q.pop();
		int l = U.L, r = U.R, cnt = U.Cnt, flag = U.Flag;
		vis[l][r][cnt][flag] = false;
		if (cnt == k) continue;
		int u;
		if (flag) u = r;
		else u = l;
		for (int ki = p[u]; ki; ki = nxt[ki]) 
			int v = to[ki], c = cost[ki];
			if (v < l) 
				if (dp[l][r][cnt][flag] + c <= dp[v][r][cnt + 1][0]) 
					dp[v][r][cnt + 1][0] = dp[l][r][cnt][flag] + c;
					if (!vis[v][r][cnt + 1][0]) 
						vis[v][r][cnt + 1][0] = true;
						q.push( v,r,cnt + 1,0 );
					
				
			
			if (v > r) 
				if (dp[l][r][cnt][flag] + c <= dp[l][v][cnt + 1][1]) 
					dp[l][v][cnt + 1][1] = dp[l][r][cnt][flag] + c;
					if (!vis[l][v][cnt + 1][1]) 
						vis[l][v][cnt + 1][1] = true;
						q.push( l,v,cnt + 1,1 );
					
				
			
		
	
	ffor(i, 1, n) 
		ffor(j, 1, n) 
			ans = min(ans, dp[i][j][k][0]);
			ans = min(ans, dp[i][j][k][1]);
		
	
	if (ans == inf) ans = -1;
	cout << ans;


int main()

	ready();
	work();
	return 0;

记忆化搜索

// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm> 
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
//#define int long long

using namespace std;

const int N = 103, M = 2020;
int n, T, k, m;
int pi, p[N], nxt[M], to[M], cost[M];

inline void add_in(int u, int v, int c) 
	pi++; nxt[pi] = p[u]; p[u] = pi; to[pi] = v; cost[pi] = c;


void ready()

	IOS;
	cin >> n >> k >> m;
	//ffor(i, 1, n) ffor(j, 1, n) ma[i][j] = INF;
	ffor(i, 1, m) 
		int u, v, c;
		cin >> u >> v >> c;
		add_in(v, u, c);
		//ma[u][v] = min(ma[u][v], c);
	


int dp[N][N][N][3];

int dfs(int l, int r, int cnt, int flag)

	if (dp[l][r][cnt][flag] != inf) return dp[l][r][cnt][flag];
	if (cnt == k) 
		return dp[l][r][cnt][flag]=0;
	
	if (flag == 0) 
		int u = l;
		for (int ki = p[u]; ki; ki = nxt[ki]) 
			int v = to[ki], c = cost[ki];
			if (v<l) 
				dp[l][r][cnt][flag] = min(dp[l][r][cnt][flag], dfs(v, r, cnt + 1, 0) + c);
			
			if (v > r) 
				dp[l][r][cnt][flag] = min(dp[l][r][cnt][flag], dfs(l, v, cnt + 1, 1) + c);
			
		
	
	else 
		int u = r;
		for (int ki = p[u]; ki; ki = nxt[ki]) 
			int v = to[ki], c = cost[ki];
			if (v < l) 
				dp[l][r][cnt][flag] = min(dp[l][r][cnt][flag], dfs(v, r, cnt + 1, 0) + c);
			
			if (v > r) 
				dp[l][r][cnt][flag] = min(dp[l][r][cnt][flag], dfs(l, v, cnt + 1, 1) + c);
			
		
	
	return dp[l][r][cnt][flag];


void work()

	mst(dp, inf);
	int ans = inf;
	ffor(i, 1, n) 
		ans = min(ans, dfs(i, i, 1, 0));
		ans = min(ans, dfs(i, i, 1, 1));
	
	if (ans == inf) ans = -1;
	cout << ans;


int main()

	ready();
	work();
	return 0;