poj 1764 Dice Contest

题目戳这里。

首先我要吐槽这个题目描述不清。(2)对着选手，那选手朝那边？看完别人写的程序后我才知道选手对着目标所在的方向（或左或右）。

然后这道题还是不错的，因为他交给我矩阵乘法不只有常规意义下的矩阵乘法，只要满足结合律，取 (min) 都行。涨姿势了。

这题我们这么做，我们用(f[i][j][k])表示骰子从起始位置到((i,j))且最后状态为(k)所需要最小代价，类似于最短路。然后直接求肯定gg，肯定是要用矩阵乘法来加速。

考虑到(y)的范围只有(4)，我们可以将后两维压成一维，就设为(g[i][j])吧。如果我们能够知道第(i)列的所有状态到第(i+1)列的所有状态的最短路，(设第((i,j))到((i+1,k))的最小代价为(h[1][j][k]))，那么我们就可以知道第(i)列所有状态到(i+n(n>1))列所有状态的最小代价了(设第((i,j))到((i+n,k))的最小代价为(h[n][j][k]))。则

[h[n][i][j] = min { h[n-1][i][k]+h[1][k][j] } ]

然后这个式子很像矩阵乘法的式子(把和改成了(min))，然后满足结合律，故也可用矩阵乘法来加速。

那么怎么求(h[1][i][j])呢？我们可以将列限制在一个范围内，然后跑最短路就行了。

这个题目最坑爹的地方就是状态处理，这个可以参见std。

#include<queue>
#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;

typedef long long ll;
const int Ex = 20,maxn = Ex*96*5; const ll inf = 1LL<<60;
const int help[6][6] = {{-1,2,4,1,3,-1},{3,-1,0,5,-1,2},{1,5,-1,-1,0,4},{4,0,-1,-1,5,1},{2,-1,5,0,-1,3},{-1,3,1,4,2,-1}};
int L[6],X1,Y1,X2,Y2,side[maxn],toit[maxn],next[maxn],len[maxn],cnt; ll dis[Ex*97],f[100],g[100][100]; bool in[Ex*97];

struct Matrix
{
	int N,M; ll S[96][96];
	inline Matrix(int n = 0,int m = 0,bool sign = false):N(n),M(m)
	{
		for (int i = 0;i < n;++i) for (int j = 0;j < m;++j) S[i][j] = inf;
		if (sign) for (int i = 0;i < n;++i) S[i][i] = 0;
	}
	friend inline Matrix operator*(const Matrix &a,const Matrix &b)
	{
		Matrix c(a.N,b.M);
		for (int i = 0;i < c.N;++i)
			for (int j = 0;j < c.M;++j)
				for (int k = 0;k < a.M;++k)
					c.S[i][j] = min(c.S[i][j],a.S[i][k]+b.S[k][j]);
		return c;
	}
};

inline void add(int a,int b,int c) { next[++cnt] = side[a]; side[a] = cnt; toit[cnt] = b; len[cnt] = c; }

inline int getid(int x,int y,int top,int front)
{
	int front_id = -1;
	for (int i = 0;i < 6;++i)
	{
		if (i != top&&i != 5-top) ++front_id;
		if (i == front) return x*96+y*24+top*4+front_id;
	}
	return -1;
}

inline void ready()
{
	for (int x = 0;x < Ex;++x)
		for (int y = 0;y < 4;++y)
			for (int top = 0;top < 6;++top)
				for (int front = 0;front < 6;++front)
				{
					if (front == top||front == 5-top) continue;
					int left = help[top][front],u = getid(x,y,top,front);
					int nx,ny,ntop,nfront,v;
					if (x)
					{
						nx = x-1; ny = y; ntop = 5-left; nfront = front;
						v = getid(nx,ny,ntop,nfront); add(u,v,L[ntop]);
					}
					if (x+1 < Ex)
					{
						nx = x+1; ny = y; ntop = left; nfront = front;
						v = getid(nx,ny,ntop,nfront); add(u,v,L[ntop]);
					}
					if (y)
					{
						nx = x; ny = y-1; ntop = 5-front; nfront = top;
						v = getid(nx,ny,ntop,nfront); add(u,v,L[ntop]);
					}
					if (y < 3)
					{
						nx = x; ny = y+1; ntop = front; nfront = 5-top;
						v = getid(nx,ny,ntop,nfront); add(u,v,L[ntop]);
					}
				}
}

inline void spfa(int source)
{
	queue <int> team;
	for (int i = 0;i < Ex*96;++i) dis[i] = inf;
	dis[source] = 0; in[source] = true; team.push(source);
	while (!team.empty())
	{
		int now = team.front(); team.pop();
		for (int i = side[now];i;i = next[i])
		{
			if (dis[toit[i]] <= dis[now]+len[i]) continue;
			dis[toit[i]] = dis[now]+len[i];
			if (!in[toit[i]]) team.push(toit[i]),in[toit[i]] = true;
		}
		in[now] = false;
	}
}

inline void record(int x,ll A[])
{
	for (int y = 0;y < 4;++y)
		for (int top = 0;top < 6;++top)
			for (int front = 0;front < 6;++front)
			{
				if (front == top||front == 5-top) continue;
				int id = getid(x,y,top,front); A[id%96] = dis[id];
			}
}

inline Matrix qsm(Matrix a,int b)
{
	Matrix ret(96,96,true);
	for (;b;b >>= 1,a = a*a) if (b&1) ret = ret*a;
	return ret;
}

inline ll work()
{
	int diff = abs(X1-X2),x = Ex >> 1; ll ans = inf;
	spfa(getid(x,Y1,0,1));
	if (!diff)
	{
		for (int top = 0;top < 6;++top)
			for (int front = 0;front < 6;++front)
			{
				if (front == top||front == 5-top) continue;
				ans = min(ans,dis[getid(x,Y2,top,front)]);
			}
		return ans;
	}
	record(x,f);
	for (int y = 0;y < 4;++y)
		for (int top = 0;top < 6;++top)
			for (int front = 0;front < 6;++front)
			{
				if (front == top||front == 5-top) continue;
				int source = getid(x,y,top,front);
				spfa(source); record(x+1,g[source%96]);
			}
	Matrix A(96,96);
	for (int i = 0;i < 96;++i) for (int j = 0;j < 96;++j) A.S[i][j] = g[i][j];
	A = qsm(A,diff);
	for (int i = 0;i < 96;++i)
		for (int j = 0;j < 96;++j)
			if (j/24 == Y2) ans = min(ans,f[i]+A.S[i][j]);
	return ans;
}

int main()
{
	freopen("1764.in","r",stdin);
	freopen("1764.out","w",stdout);
	for (int i = 0;i < 6;++i) scanf("%d",L+i);
	scanf("%d %d %d %d",&X1,&Y1,&X2,&Y2); --Y1,--Y2;
	if (X1 > X2) swap(L[2],L[3]);
	ready();
	cout << work() << endl;
	fclose(stdin); fclose(stdout);
	return 0;
}