HDU 6395 分段矩阵快速幂 HDU 6386 建虚点+dij

http://acm.hdu.edu.cn/showproblem.php?pid=6395

Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1475    Accepted Submission(s): 539

Problem Description

Let us define a sequence as below

⎧⎩⎨⎪⎪⎪⎪⎪⎪F1F2Fn===ABC⋅Fn−2+D⋅Fn−1+⌊Pn⌋

  Your job is simple, for each task, you should output Fn module 109+7.

 

Input

The first line has only one integer T, indicates the number of tasks.

Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n.

1≤T≤200≤A,B,C,D≤1091≤P,n≤109

 

Sample Input

2 3 3 2 1 3 5 3 2 2 2 1 4

 

Sample Output

36 24

 

解析  p/n 不是定值不能直接矩阵快速幂  但是观察发现

23 

 1     2   3   4   5   6   7   8   9  10  11  12  13...100   除数

 23  12  7   5   4   3   3   2    2   2   2   1    1 .... 0      商

    商的个数很少 我们可以 分段矩阵快速幂   怕超时 可以先打1e5的表    一开始以为要用逆元解决问题  真的智障了

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("
");
#define debug(a,b) cout<<a<<" "<<b<<" "<<endl;
using namespace std;
typedef long long ll;
const ll maxn=3,inf=0x3f3f3f3f,mod=1e9+7;
bool Finish_read;
template<class T>inline void read(T &x)
{
    Finish_read=0;
    x=0;
    int f=1;
    char ch=getchar();
    while(!isdigit(ch))
    {
        if(ch=='-')f=-1;
        if(ch==EOF)return;
        ch=getchar();
    }
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
    Finish_read=1;
}
template<class T>inline void print(T x)
{
    if(x/10!=0)print(x/10);
    putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    if(x<0)putchar('-');
    x=abs(x);
    print(x);
    putchar('
');
}
template<class T>inline void write(T x)
{
    if(x<0)putchar('-');
    x=abs(x);
    print(x);
}
ll n,a,b,c,d,p;
struct Matrix
{
    ll m[maxn][maxn];
    Matrix()
    {
        memset(m,0,sizeof(m));
    }
    void init()
    {
        for(int i=0; i<maxn; i++)
            for(int j=0; j<maxn; j++)
                m[i][j]=(i==j);
    }
    Matrix  operator +(const Matrix &b)const
    {
        Matrix c;
        for(int i=0; i<maxn; i++)
        {
            for(int j=0; j<maxn; j++)
            {
                c.m[i][j]=(m[i][j]+b.m[i][j])%mod;
            }
        }
        return c;
    }
    Matrix  operator *(const Matrix &b)const
    {
        Matrix c;
        for(int i=0; i<maxn; i++)
        {
            for(int j=0; j<maxn; j++)
            {
                for(int k=0; k<maxn; k++)
                {
                    c.m[i][j]=(c.m[i][j]+(m[i][k]*b.m[k][j])%mod)%mod;
                }
            }
        }
        return c;
    }
    Matrix  operator^(const ll &t)const
    {
        Matrix ans,a=(*this);
        ans.init();
        ll n=t;
        while(n)
        {
            if(n&1) ans=ans*a;
            a=a*a;
            n>>=1;
        }
        return ans;
    }
};
ll f[100010];
pair<ll,ll> solve(ll x,ll y,ll z,ll m)
{
    Matrix a,b,temp;
    b.m[0][0]=d;b.m[0][1]=c;b.m[0][2]=1;
    b.m[1][0]=1;b.m[2][2]=1;
    a.m[0][0]=x;
    a.m[1][0]=y;
    a.m[2][0]=z;
    temp=b^(m);
    temp=temp*a;
    return mp(temp.m[0][0],temp.m[1][0]);
}
int main()
{
    int t;
    read(t);
    while(t--)
    {
        read(a);read(b);read(c);read(d);read(p);read(n);
        f[1]=a;f[2]=b;
        for(int i=3;i<=100000;i++)
            f[i]=(c*f[i-2]%mod+d*f[i-1]%mod+p/i)%mod;
        if(n<=100000)
            writeln(f[n]);
        else
        {
            ll l=100000;
            pair<ll,ll> pp(f[l],f[l-1]);l++;
            while(l<=n)
            {
                ll temp=p/l;
                if(temp==0)
                {
                    pp=solve(pp.fi,pp.se,temp,n-l+1);
                    writeln(pp.fi);
                    break;
                }
                ll temp2=p/temp;
                if(temp2>=n)
                {
                    pp=solve(pp.fi,pp.se,temp,n-l+1);
                    writeln(pp.fi);
                    break;
                }
                else
                {
                    pp=solve(pp.fi,pp.se,temp,temp2-l+1);
                    l=temp2+1;
                }
            }
        }
    }
}

http://acm.hdu.edu.cn/showproblem.php?pid=6386

Age of Moyu

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2237    Accepted Submission(s): 683

Problem Description

Mr.Quin love fishes so much and Mr.Quin’s city has a nautical system,consisiting of N ports and M shipping lines. The ports are numbered 1 to N. Each line is occupied by a Weitian. Each Weitian has an identification number.

The i-th (1≤i≤M) line connects port Ai and Bi (Ai≠Bi) bidirectionally, and occupied by Ci Weitian (At most one line between two ports).

When Mr.Quin only uses lines that are occupied by the same Weitian, the cost is 1 XiangXiangJi. Whenever Mr.Quin changes to a line that is occupied by a different Weitian from the current line, Mr.Quin is charged an additional cost of 1 XiangXiangJi. In a case where Mr.Quin changed from some Weitian A's line to another Weitian's line changes to Weitian A's line again, the additional cost is incurred again.

Mr.Quin is now at port 1 and wants to travel to port N where live many fishes. Find the minimum required XiangXiangJi (If Mr.Quin can’t travel to port N, print −1instead)

 

Input

There might be multiple test cases, no more than 20. You need to read till the end of input.

For each test case,In the first line, two integers N (2≤N≤100000) and M (0≤M≤200000), representing the number of ports and shipping lines in the city.

In the following m lines, each contain three integers, the first and second representing two ends Ai and Bi of a shipping line (1≤Ai,Bi≤N) and the third representing the identification number Ci (1≤Ci≤1000000) of Weitian who occupies this shipping line.

 

Output

For each test case output the minimum required cost. If Mr.Quin can’t travel to port N, output −1 instead.

 

Sample Input

3 3 1 2 1 1 3 2 2 3 1 2 0 3 2 1 2 1 2 3 2

 

Sample Output

1 -1 2

 

解析 这题标程有些错误 而且是原题 既然标程是错的 也就不用按照 题解来写了 感觉现在最靠谱的思路就是 建虚点了 边权相同且相邻的点 建立一个虚点x

其他点到x距离为1的单向边 x到其他点距离为0单向边 把原来的去掉 跑最短路就是答案。这道题真是服了。。。

#include <bits/stdc++.h>
#define Pii pair<int,int>
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
const int maxm=2e5+5;
const int maxnn=1e6+5;
const int inf=2147483647;
vector<int> GG[maxn];
int from[maxm],to[maxm],c[maxm];
int n,cnt,nowc,dis[maxnn];             //cnt:n+虚点数
bool v[maxm],vis[maxnn];
struct node{
    int d,x;
    bool operator < (const node& b) const
    {return d>b.d;}
};
priority_queue<node> Q;
int _first[maxnn],_tip[maxnn],_w[maxnn],_next[maxnn],edge;
inline void dijk()
{
    int i,u,vv,len,num;
    node tmp;
    for(i=1;i<=cnt;i++) dis[i]=inf;
    dis[1]=0;
    Q.push((node){0,1});
    while(!Q.empty())
    {
        tmp=Q.top();
        Q.pop();
        u=tmp.x;
        if(vis[u]==true) continue;
        vis[u]=true;
        num=_first[u];
        while(num!=-1)
        {
            vv=_tip[num];
            if(dis[u]!=inf&&dis[vv]>dis[u]+_w[num])
            {
                dis[vv]=dis[u]+_w[num];
                Q.push((node){dis[vv],vv});
            }
            num=_next[num];
        }
    }
    return;
}
inline void dfs(int x)
{
    //add(x,cnt,1);
    _tip[++edge]=cnt;
    _w[edge]=1;
    _next[edge]=_first[x];
    _first[x]=edge;

    //add(cnt,x,0);
    _tip[++edge]=x;
    _w[edge]=0;
    _next[edge]=_first[cnt];
    _first[cnt]=edge;

    int i,len=GG[x].size(),num;
    for(i=0;i<len;i++)
    {
        num=GG[x][i];
        if(c[num]==nowc&&!v[num])
        {
            v[num]=true;
            if(from[num]==x) dfs(to[num]);
            else dfs(from[num]);
        }
    }
    return;
}
int main()
{
    int i,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        edge=0;
        cnt=n;
        memset(v,0,sizeof(v));
        memset(vis,0,sizeof(vis));
        memset(_first,-1,sizeof(_first));
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%d",&from[i],&to[i],&c[i]);
            GG[from[i]].push_back(i);
            GG[to[i]].push_back(i);
        }
        for(i=1;i<=m;i++)
            if(!v[i])
            {
                cnt++;
                nowc=c[i];
                dfs(from[i]);
            }
        dijk();
        if(dis[n]==inf) printf("-1
");
        else printf("%d
",dis[n]);
        //test();
        for(i=1;i<=n;i++) GG[i].clear();
    }
    return 0;
}

转自  https://blog.csdn.net/jerry99s/article/details/81664117