poj3613（矩陣快速冪+最短路）

題目鏈接：http://poj.org/problem?id=3613

思路：T很小，所以很容易想到Floyd，但是N很大，直接跑肯定超時，這個轉移狀態滿足結合律，所以可以跑矩陣快速冪，自乘一次代表多走一條邊（代碼poj過不了編譯的，因爲我把頭文件什麼的又改了回去）

#pragma GCC optimize(2)
#include <bits/stdc++.h>
#include <ext/pb_ds/priority_queue.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
#include <iostream>
using namespace std;
const int N=2e5+8;
int n,x,y,NN,T,S,E,z;
unordered_map<int,int>mp;
struct matrix{
ll a[105][108];};
matrix base;
matrix mul(matrix x,matrix y)
{
    matrix res;
    REW(res.a,inff);
    FOR(i,1,n)
     FOR(j,1,n)
      FOR(k,1,n) res.a[i][j]=min(res.a[i][j],x.a[i][k]+y.a[k][j]);
     return res;
}
matrix gxmod(matrix x,ll b)
{
    matrix res=x;
    while(b)
    {
        if(b&1) res=mul(res,x);
        x=mul(x,x),b>>=1;
    }
    return res;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>NN>>T>>S>>E;
    REW(base.a,inff);
    while(T--)
    {
        cin>>z>>x>>y;
        x=mp.count(x)?mp[x]:mp[x]=++n;
        y=mp.count(y)?mp[y]:mp[y]=++n;
        base.a[x][y]=base.a[y][x]=z;
    }
    matrix ans=gxmod(base,NN-1);
    cout<<ans.a[mp[S]][mp[E]]<<endl;
    return 0;
}

還有洛谷的P6190也是一樣類型的題

題目鏈接：https://www.luogu.com.cn/problem/P6190

#pragma GCC optimize(2)
#include <bits/stdc++.h>
#include <ext/pb_ds/priority_queue.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
#include <iostream>
using namespace std;
const int N=3e3+8;
ll n,m,k,x[N],y[N],z[N],a[N][N];
struct matrix{
ll a[108][108];};
matrix base;
matrix mul(matrix x,matrix y)
{
    matrix res;
    REW(res.a,inff);
    FOR(i,1,n)
     FOR(j,1,n)
      FOR(k,1,n) res.a[i][j]=min(res.a[i][j],x.a[i][k]+y.a[k][j]);
     return res;
}
matrix gxmod(matrix x,ll b)
{
    matrix res=x;
    while(b)
    {
        if(b&1) res=mul(res,x);
        x=mul(x,x),b>>=1;
    }
    return res;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n>>m>>k;
    REW(base.a,inff);
    FOR(i,1,m)
    {
        cin>>x[i]>>y[i]>>z[i];
        base.a[x[i]][y[i]]=z[i];
    }
    FOR(i,1,n) base.a[i][i]=0;
    FOR(k,1,n)
     FOR(i,1,n)
      FOR(j,1,n) a[i][j]=base.a[i][j]=min(base.a[i][j],base.a[i][k]+base.a[k][j]);
    if(!k) return cout<<base.a[1][n]<<endl,0; 
    FOR(k,1,m)
     FOR(i,1,n)
      FOR(j,1,n) base.a[i][j]=min(base.a[i][j],a[i][x[k]]+a[y[k]][j]-z[k]);
    matrix ans=gxmod(base,k-1);
    cout<<ans.a[1][n]<<endl;
    return 0;
}