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赛时靠spfa求最长路骗了30pts

• spfa的时间复杂度是\(O(k|E|)\)，不是\(O(k|N|)\)！
• dijkstra 时间复杂度\(O((n+m)logn)\)
• 特别注意这两个的复杂度都和边数密切相关

spfa的话按a值分层，按层建边即可

正解是个dp，考场上想到dp，但dp思路错了

令\(dp[i][j]\)为在\((i, j)\)位置结束时的最大和

则

\[dp[i][j] = max(b[i][j]+dp[i'][j']+|i-i'|+|j-j'|)
\]

又一个带绝对值的题，可以分情况讨论

对于一个点\((i, j)\)，可以用四个二维树状数组分别维护\(i,j\)为正，负时的最大值，

根据点\((i, j)\)和\((i', j')\)的位置关系查询即可，复杂度\(O(nmlognlogm)\)

但是还有一种\(O(nm)\)的解法：

• 常用距离算法详解

题面里那个 \(|i-i'|+|j-j'|\) 其实是曼哈顿距离

这里我们要求其最大值

• 曼哈顿距离在处理绝对值时，常转换为切比雪夫距离以简化计算

考虑转化为切比雪夫距离，那方程可以化为

\[dp[i][j] = b[i][j]+max(dp[i'][j']+max(|i-i'|,|j-j'|))
\]

要求距离尽可能大，所以我们可以维护全局 \(max\ \{dp[i'][j']+i', dp[i'][j']-i', dp[i'][j']+j', dp[i'][j']-j'\}\)， 转移时用这个分情况转移

yysy，我因为脑残没仔细看题面a=0的点舍弃从下午3点调到6点

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x7fffffffffff
#define N 2010
#define ll long long
#define ld long double
#define usd unsigned
#define ull unsigned long long
//#define int long long 

#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
char buf[1<<21], *p1=buf, *p2=buf;
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m;

namespace force{
	int tot, pos;
	int head[N*N], size, lim[N*N], lcnt; ll dis[N*N];
	bool vis[N*N];
	int ss, st;
	struct point{int a, b, x, y, rk; ll w;}mp[N*N];
	inline bool operator < (point a, point b) {return a.a<b.a;}
	struct edge{int to, next, val;}e[N*N*3];
	inline void add(int s, int t, int w) {edge* k=&e[++size]; k->to=t; k->val=w; k->next=head[s]; head[s]=size;}
	void spfa(int s) {
		queue<int> q;
		dis[s]=0; vis[s]=1; q.push(s);
		int t;
		while (q.size()) {
			t=q.front(); q.pop();
			vis[t]=0;
			for (int i=head[t],v; i; i=e[i].next) {
				v = e[i].to;
				//cout<<t<<": "<<dis[t]<<' '<<mp[t].w<<' '<<e[i].val<<' '<<dis[v]<<endl;
				if (dis[t]+mp[t].w+e[i].val < dis[v]) {
					dis[v] = dis[t]+mp[t].w+e[i].val;
					if (!vis[v]) q.push(v);
				}
			}
		}
		//cout<<"dis: "; for (int i=1; i<=tot+2; ++i) cout<<dis[i]<<' '; cout<<endl;
	}
	void solve() {
		tot=0; for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) mp[++tot].a=read(), mp[tot].x=i, mp[tot].y=j, mp[tot].rk=tot;
		tot=0; for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) mp[++tot].w=-read();
		sort(mp+1, mp+tot+1);
		//for (int i=1; i<=tot; ++i) cout<<mp[i].a<<' '; cout<<endl;
		for (pos=1; pos<=tot; ++pos) if (mp[pos].a!=mp[pos-1].a) lim[++lcnt]=pos;
		lim[++lcnt]=tot+1;
		for (int i=1; i<=tot+10; ++i) dis[i]=INF;
		ss=tot+1; st=tot+2;
		for (int i=lim[1]; i<lim[2]; ++i) add(ss, i, 0); //, cout<<"ss: "<<ss<<' '<<i<<endl;
		for (int i=lim[lcnt-1]; i<=tot; ++i) add(i, st, 0); //, cout<<"st: "<<i<<' '<<st<<endl;
		for (int i=1; i<lcnt; ++i) {
			for (int j=lim[i]; j<lim[i+1]; ++j) {
				for (int k=lim[i+1]; k<lim[i+2]; ++k) add(j, k, -(abs(mp[j].x-mp[k].x)+abs(mp[j].y-mp[k].y))); //, cout<<"add "<<j<<' '<<k<<' '<<-(abs(mp[j].x-mp[k].x)+abs(mp[j].y-mp[k].y))<<endl;
			}
		}
		spfa(ss);
		printf("%lld\n", -dis[st]);
	}
}

namespace task{
	// 0->i 1->-i 2->j 3->-j
	int tot, lcnt, pos[4], pos2;
	ll maxn[4], dp[N*N], ans;
	struct point{int a, i, j; ll b;}p[N*N];
	inline bool operator < (point a, point b) {return a.a<b.a;}
	struct line{int l, r; inline void build(int l_, int r_) {l=l_; r=r_;}}lin[N*N];
	void solve() {
		memset(maxn, 128, sizeof(maxn));
		tot=0; for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) p[++tot].a=read(), p[tot].i=i+j, p[tot].j=i-j;
		tot=0; for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) p[++tot].b=read();
		sort(p+1, p+tot+1);
		for (pos2=1; !p[pos2].a; ++pos2) ;
		for (int i=pos2; i<=tot; ++i) {
			lin[++lcnt].l=i;
			for (int j=i+1; j<=tot&&(p[j].a==p[j-1].a); ++i,++j) ;
			lin[lcnt].r=i;
		}
		//for (int i=1; i<=lcnt; ++i) cout<<lin[i].l<<","<<lin[i].r<<endl;
		//for (int i=1; i<=tot; ++i) cout<<p[i].i<<' '<<p[i].j<<endl;
		int i1, j1;
		for (int j=lin[1].l; j<=lin[1].r; ++j) {
			i1=p[j].i, j1=p[j].j;
			dp[j] = p[j].b;
			ans = max(ans, dp[j]);
		}
		for (int j=lin[1].l; j<=lin[1].r; ++j) {
			i1=p[j].i, j1=p[j].j;
			//cout<<"ij: "<<i1<<' '<<j1<<endl;
			maxn[0] = max(maxn[0], dp[j]+i1);
			maxn[1] = max(maxn[1], dp[j]-i1);
			maxn[2] = max(maxn[2], dp[j]+j1);
			maxn[3] = max(maxn[3], dp[j]-j1);
		}
		//cout<<"ans: "<<ans<<endl;
		//cout<<"maxn: "; for (int i=0; i<4; ++i) cout<<maxn[i]<<' '; cout<<endl;
		for (int i=2; i<=lcnt; ++i) {
			for (int j=lin[i].l; j<=lin[i].r; ++j) {
				i1=p[j].i, j1=p[j].j;
				dp[j] = p[j].b+max(max(max(maxn[0]-i1, maxn[1]+i1), maxn[2]-j1), maxn[3]+j1);
				//cout<<"ans in "<<maxn[0]-i1<<' '<<maxn[1]+i1<<' '<<maxn[2]-j1<<' '<<maxn[3]+j1<<endl;
				ans = max(ans, dp[j]);
			}
			for (int j=lin[i].l; j<=lin[i].r; ++j) {
				i1=p[j].i, j1=p[j].j;
				maxn[0] = max(maxn[0], dp[j]+i1);
				maxn[1] = max(maxn[1], dp[j]-i1);
				maxn[2] = max(maxn[2], dp[j]+j1);
				maxn[3] = max(maxn[3], dp[j]-j1);
			}
			//cout<<"ans: "<<ans<<endl;
			//cout<<"maxn: "; for (int i=0; i<4; ++i) cout<<maxn[i]<<' '; cout<<endl;
		}
		printf("%lld\n", ans);
	}
}

signed main()
{
	#ifdef DEBUG
	freopen("1.in", "r", stdin);
	#endif

	n=read(); m=read();
	task::solve();

	return 0;
}