题意
给定一个有向图,每条边的花费为1。现在有一个空间跑路器,可以走2^k长度的路,只用花1秒的时间。问从1走到n最少的时间。
n <= 50, k <= 64。
思路
这道题说是倍增,但是写起来不见倍增的影子,我觉得真妙,对倍增有了更膜拜的认识。
我们可以开一个bool矩阵dp【i】【j】【k】,表示i到j是否可以通过2^k的路程到达。更新这个矩阵可以通过类似floyd最短路的思想
if(dp[i][t][k-1] && dp[t][j][k-1]) dp[i][j][k] = 1;
再开一个dis【i】【j】记录距离,跑一遍floyd就可以了
复杂度是O(n*n*n*64)
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> /* ⊂_ヽ \\ Λ_Λ 来了老弟 \(‘?‘) > ⌒ヽ / へ\ / / \\ ? ノ ヽ_つ / / / /| ( (ヽ | |、\ | 丿 \ ⌒) | | ) / ‘ノ ) L? */ using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ‘\n‘ #define boost ios::sync_with_stdio(false);cin.tie(0) #define rep(a, b, c) for(int a = (b); a <= (c); ++ a) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar(); while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar(); return x=f?-x:x; } inline void cmax(int &x,int y){if(x<y)x=y;} inline void cmax(ll &x,ll y){if(x<y)x=y;} inline void cmin(int &x,int y){if(x>y)x=y;} inline void cmin(ll &x,ll y){if(x>y)x=y;} /*-----------------------showtime----------------------*/ const int maxn = 109; int n,m; int dp[maxn][maxn][maxn]; ll dis[maxn][maxn]; int main(){ scanf("%d%d", &n, &m); rep(i, 1, n) rep(j, 1, n) dis[i][j] = inff; for(int i=1; i<=m; i++) { int x,y; scanf("%d%d", &x, &y); dis[x][y] = 1; dp[x][y][0] =1; } for(int k=1; k<=64; k++){ for(int i=1; i<=n; i++){ for(int j=1; j<=n; j++){ for(int t=1; t<=n; t++){ if(dp[i][t][k-1] && dp[t][j][k-1]){ dp[i][j][k] = 1; dis[i][j] = 1; } } } } } for(int i=1; i<=n; i++){ for(int j=1; j<=n; j++){ for(int k=1; k<=n; k++){ if(dis[i][j] > dis[i][k] + dis[k][j]) dis[i][j] = dis[i][k] + dis[k][j]; } } } printf("%lld\n", dis[1][n]); return 0; }
原文地址:https://www.cnblogs.com/ckxkexing/p/10381288.html
时间: 2024-11-05 23:18:57