poj 1764 Dice Contest

最新推荐文章于 2020-12-20 01:53:12 发布

转载最新推荐文章于 2020-12-20 01:53:12 发布 · 152 阅读

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原文链接：http://www.cnblogs.com/mmlz/p/6363846.html

本文介绍了一种使用矩阵乘法优化动态规划的方法解决特定路径寻找问题。通过将传统意义上的矩阵乘法替换为求最小值操作，实现了路径寻找问题的有效解决。

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;
}

转载于:https://www.cnblogs.com/mmlz/p/6363846.html