https://www.luogu.org/problemnew/show/P1613

題意：每一秒鐘可以跑千米，求最短路

C++版本一

題解：倍增+最短路（Dijkstra）

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[33][N][N]; bool vis[N]; char str; struct node{int u,v,w;node(){};node(int _u,int _v,int _w):u(_u),v(_v),w(_w){}bool operator < (const node &S)const{return w>S.w;} }x,tmp; vector<node>G[N]; priority_queue<node>q; void ST(){for(int p=0;p<=31;p++){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){for(int k=1;k<=n;k++){a[p+1][i][j]=a[p+1][i][j]||(a[p][i][k]&&a[p][k][j]);if(a[p+1][i][j]){G[i].push_back({i,j,1});//cout<<p+1<<" "<<i<<" "<<j<<endl;break;}}}}} } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){scanf("%d%d",&u,&v);a[0][u][v]=1;G[u].push_back({u,v,1});}ST();q.push({0,1,0});vis[1]=1;while(!q.empty()){x=q.top();q.pop();//cout<<x.v<<endl;if(x.v==n){break;}int u=x.v;for(int i=0,j=G[u].size();i<j;i++){int v=G[u][i].v;if(!vis[v]){vis[v]=1;q.push({0,v,x.w+1});}}}cout<<x.w<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

C++版本二

題解：倍增+最短路（Floyd）

#include<bits/stdc++.h> using namespace std; const int N = 110; int n,m; int u,v; int a[N][N]; bool f[N][N][N]; int main() {memset(a,0x3f3f3f,sizeof(a));cin>>n>>m;for(int i=1;i<=n;i++)a[i][i]=0;for(int i=1;i<=m;i++){cin>>u>>v;f[u][v][0]=1;if(u==v)a[u][v]=0;else a[u][v]=1;}for(int s=1;s<=20;s++)for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(f[i][k][s-1] && f[k][j][s-1]){f[i][j][s]=1;if(i==j)a[i][j]=0;else a[i][j]=1;}for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)a[i][j]=min(a[i][j],a[i][k]+a[k][j]);cout<<a[1][n];return 0; }

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