[插头DP自我总结]

[HNOI 2007]神奇游乐园

#include <bits/stdc++.h>
#define maxn 110
using namespace std;
typedef long long ll;
int n, m;
int a[maxn][maxn];
ll ans = -1ll << 60;
#define M 2000010
#define mod 997

struct Hashmap{

	ll st[M];
	int h[1000], size, nxt[M];
	ll f[M];
	void clear(){memset(h, 0, sizeof h); size = 0;}
	void push(ll hash_, ll val){
		int tmp = hash_ % mod;
		for(int i = h[tmp]; i; i = nxt[i]){
			if(st[i] == hash_){
				f[i] = max(f[i], val);
				return;
			}
		}
		int now = ++ size;
		f[now] = val;
		st[now] = hash_;
		nxt[now] = h[tmp];
		h[tmp] = now;
	}
}dp[2];

int cur, code[20], ch[20];

void Decode(ll st){
	for(int i = m; i >= 0; i --)
	    code[i] = st & 7, st >>= 3;
}

ll Encode(){
	ll ret = 0;
	memset(ch, -1, sizeof ch);
	ch[0] = 0; int cnt = 0;
	for(int i = 0; i <= m; i ++){
		if(ch[code[i]] == -1)ch[code[i]] = ++ cnt;
		code[i] = ch[code[i]];
		ret = ret << 3 | code[i];
	}
	return ret;
}

void Shift(){
	for(int i = m; i >= 1; i --)
	    code[i] = code[i-1];
	code[0] = 0;
}

inline void Change(int u, int v){
	for(int i = 0; i <= m; i ++)
	    if(code[i] == v)
	        code[i] = u;
}

void DP(int i, int j){
	dp[cur^1].clear();
	for(int k = 1; k <= dp[cur].size; k ++){
		Decode(dp[cur].st[k]);
		if(j == 1){if(code[m])continue;Shift();}
		int Left = code[j-1], Up = code[j];
		if(Left && Up){
			code[j] = code[j-1] = 0;
			if(Left == Up){
                ll ENCODE = Encode();
				if(ENCODE == 0)ans = max(ans, dp[cur].f[k] + a[i][j]);
			}
			else{
                Change(Left, Up);
                dp[cur^1].push(Encode(), dp[cur].f[k] + a[i][j]);
			}
		}
		else if(Left || Up){
			int tmp = Left ? Left : Up;
			code[j-1] = 0, code[j] = tmp;
			dp[cur^1].push(Encode(), dp[cur].f[k] + a[i][j]);
			code[j] = 0, code[j-1] = tmp;
			dp[cur^1].push(Encode(), dp[cur].f[k] + a[i][j]);
		}
		else{
            dp[cur^1].push(Encode(), dp[cur].f[k]);
            code[j] = code[j-1] = 8;
            dp[cur^1].push(Encode(), dp[cur].f[k] + a[i][j]);
		}
	}
	cur ^= 1;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("park.in","r",stdin);
    freopen("park.out","w",stdout);
#endif
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i ++)
		for(int j = 1; j <= m; j ++)
			scanf("%d", &a[i][j]);
	dp[cur].clear();
	dp[cur].push(0, 0);
	for(int i = 1; i <= n; i ++)
		for(int j = 1; j <= m; j ++)
		    DP(i, j);

	printf("%lld\n", ans);
	return 0;
}

粘了一个模板上来=-=

基于连通性的动态规划，最小表示法很好用。

我们可以用一个压缩的数字表示一个连通情况，比如DP生成树，概率，棋盘上格子的情况，等等等

只要和连通性有关而且n很小时就可以用啦QAQ。

又忘了模板了QAQ

就是如果左边和上面是一个连通分量即Left == Up时，我们要合并连通分量，所以此时已经出现了一个圈了，这道题不能有多个圈，所以最后不放进去

code[j] = code[j-1] = 0.

[BZOJ 3753]Wall

扩展一下方格变成一个回路问题，然后射线法判断格子是否内部。

出现的问题是当left == up时，code[j] = code[j-1] = 0，然后再改所有的标号

当必须只能有一个回路的时候，Encode()一定要等于0.

#include <bits/stdc++.h>
#define maxn 20
using namespace std;
typedef long long ll;
int n, m;

int need[maxn][maxn], a[maxn][maxn];

int cur = 0, Cnt, cnt;

const int md = 997;

struct Hashmap{
	#define M 2000010
	ll st[M];
	int f[M], h[1000], nxt[M], size;
	void init(){
		memset(h, 0, sizeof h);
		size = 0;
	}
	void push(ll hs, int val){
		int tmp = hs % md;
		for(int i = h[tmp]; i; i = nxt[i]){
			if(st[i] == hs){
				f[i] = max(f[i], val);
				return;
			}
		}
		int now = ++ size;
		f[now] = val;
		nxt[now] = h[tmp];
		st[now] = hs;
		h[tmp] = now;
	}
}dp[2];

int code[maxn], ch[maxn];

void Decode(ll st){
	for(int i = m; i >= 0; i --)
	    code[i] = st & 7, st >>= 3;
}

ll Encode(){
	ll ret = 0;
	memset(ch, -1, sizeof ch);
	ch[0] = 0; int cnt = 0;
	for(int i = 0; i <= m; i ++){
		if(ch[code[i]] == -1)
		    ch[code[i]] = ++ cnt;
		code[i] = ch[code[i]];
		ret <<= 3;
		ret |= code[i];
	}return ret;
}

void Shift(){
	for(int i = m; i; i --)
	    code[i] = code[i-1];
	code[0] = 0;
}

bool Judge(int cnt, int i, int j){
	if(cnt && need[i][j] == 1)return false;
	if(!cnt && need[i][j] == 2)return false;
	return true;
}

int ans = -0x7fffffff;

void Trans(int i, int j){
	dp[cur^1].init();
	for(int k = 1; k <= dp[cur].size; k ++){
		Decode(dp[cur].st[k]);
		if(j == 1){if(code[m])continue;Shift();}
		int lf = code[j-1], up = code[j], sta = 0;
		for(int p = 0; p < j-1; p ++)
		    sta ^= (code[p] != 0);
		if(lf && up){
            code[j] = code[j-1] = 0;
			if(lf == up){
				if(Encode() == 0 && cnt == Cnt)//here!
					ans = max(ans, dp[cur].f[k]);
			}
			else{
				if(Judge(sta, i, j)){
					for(int p = 0; p <= m; p ++)
					    if(code[p] == up)
					        code[p] = lf;
					dp[cur^1].push(Encode(), dp[cur].f[k] + sta * a[i][j]);
				}
			}
		}
		else if(lf || up){
			int tmp = lf ? lf : up;
			if(Judge(sta, i, j)){
                code[j-1] = 0, code[j] = tmp;
                dp[cur^1].push(Encode(), dp[cur].f[k] + sta * a[i][j]);
			}
			sta ^= 1;
			if(Judge(sta, i, j)){
				code[j] = 0, code[j-1] = tmp;
				dp[cur^1].push(Encode(), dp[cur].f[k] + sta * a[i][j]);
			}
		}
		else{
			if(Judge(sta, i, j))
				dp[cur^1].push(Encode(), dp[cur].f[k] + sta * a[i][j]);
			sta ^= 1;
            if(Judge(sta, i, j)){
				code[j] = code[j-1] = 15;
				dp[cur^1].push(Encode(), dp[cur].f[k] + sta * a[i][j]);
            }
		}
	}
	if(need[i][j] == 2)cnt ++;
	cur ^= 1;
}

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i ++)
		for(int j = 1; j <= m; j ++)
		    scanf("%d", &a[i][j]);
    n ++, m ++;
    dp[cur].init();
    dp[cur].push(0, 0);
    for(int i = 1; i <= n; i ++)
		for(int j = 1; j <= m; j ++)
		    Trans(i, j);
	printf("%d\n", ans);

    for(int i = 1; i < n; i ++)
		for(int j = 1; j < m; j ++)
		    scanf("%d", &need[i][j]), Cnt += need[i][j] == 2;
	ans = -0x7fffffff;
	dp[cur].init();
	dp[cur].push(0, 0);
	for(int i = 1; i <= n; i ++)
		for(int j = 1; j <= m; j ++)
		    Trans(i, j);
	if(ans < -0x7ffffff)printf("Can not establish GFW.");
	else printf("%d\n", ans);
	return 0;
}