P1613 跑路(倍增 + floyd)

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

思路:

1.读入

2.建图

3.对于每一个点,向距离它 2^k 长度的点连一条长度为 1 的边

4.在新图上跑1~n的最短路

实现

看题解后,发现思路大致正确,但实现有问题;

具体看代码

 #include <bits/stdc++.h>
 #define read read()
 #define up(i,l,r) for(register int i = (l);i <= (r);i++)
 #define down(i,l,r) for(register int i = (l);i >= (r);i--)
 #define traversal_vedge(i) for(register int i = head[u]; i ;i = e[i].nxt)
 #define ll long long
 using namespace std;
 int read
 {
     int x = , f = ; char ch = getchar();
     while(ch <  || ch > ) {if(ch == '-')f = -; ch = getchar();}
     while(ch >= && ch <=) {x =  * x + ch - ;ch = getchar();}
     return x * f;
 }
 void write(int x)
 {
     if(x < ) x = -x,putchar('-');
     if(x > ) write(x/);
     putchar(x% + );
 }
 //----------------------------------------------------------
 const int N = ;
 int n,m;
 int dis[N][N];
 bool G[N][N][];

 void readdata()
 {
     memset(dis,0x3f,sizeof(dis));
     n = read; m = read;
     int u,v;
     up(i,,m)
     {
         u = read; v = read;
         dis[u][v] = ;
         G[u][v][] = ;
     }
 }

 void init()
 {
     up(p,,)
         up(i,,n)
             up(k,,n)
                 up(j,,n)
                 {
                     if(G[i][k][p-] ==  && G[k][j][p-] == )
                     G[i][j][p] = ,dis[i][j] = ;//debug G[i][j][k] -> G[i][j][p]
                 }
     //预处理
 }

 void work()
 {
     up(k,,n) up(i,,n) up(j,,n) dis[i][j] = min(dis[i][j],dis[i][k] + dis[k][j]);
     //floyd
     printf("%d\n",dis[][n]);
 }

 int main()
 {
     freopen("input.txt","r",stdin);
     readdata();
     init();
     work();
     return ;
 }