The triple loop we wrote in those years -- CodeForces 295B Greg and Graph review the Floyd algorithm

Greg and Graphtime limit per test

3 seconds

Memory limit per test

256 megabytes

Input

Standard input

Output

Standard output

Greg has a weighed directed graph, consistingNVertices. In this graph any pair of distinct vertices has an edge between them in both
Directions. Greg loves playing with the graph and now he has published Ted a new game:

• The game consistsNSteps.
• OnI-Th step Greg removes vertex numberXIFrom
  The graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex.
• Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. the shortest path can go through any remaining vertex. in other words, if we assume thatD(I, Bytes,V, Bytes,U) Is
  The shortest path between verticesVAndUIn
  The graph that formed before deleting vertexXI,
  Then Greg wants to know the value of the following sum :.

Help Greg, print the value of the required sum before each step.

Input

The first line contains integerN(1 digit ≤ DigitNLimit ≤ limit 500 )-
The number of vertices in the graph.

NextNLines containNIntegers
Each-the graph adjacency matrix:J-Th number inI-Th
LineAIj(1 digit ≤ DigitAIjLimit ≤ limit 105, limit,AIiBytes = bytes 0) represents
The weight of the edge that goes from vertexITo vertexJ.

The next line containsNDistinct integers:X1, bytes,X2, middle..., middle ,...,XN(1 digit ≤ DigitXILimit ≤ limitN)-
The vertices that Greg deletes.

Output

PrintNIntegers-I-Th
Number equals the required sum beforeI-Th step.

Please, do not use the % lld specifier to read or write 64-bit integers in C ++.
It is preferred to use thecin, cout streams
Of the % I64d specifier.

Sample test (s) input

101

Output

0 

Input

20 54 01 2

Output

9 0 

Input

40 3 1 16 0 400 12 4 0 11 1 1 04 1 2 3

Output

17 23 404 0 

Link: here

This is a classic question worth doing. It reminds us of the light behind the triple for loop of the Floyd algorithm.

#include<cstdio>#define N 505#define ll long long intll dp[N][N],arr[N],ans[N],n;ll _min(ll a,ll b) {    return a<b?a:b;}int main(){    while(scanf("%d",&n)!=EOF) {        for(int i=0;i<n;i++) {            for(int j=0;j<n;j++) scanf("%lld",&dp[i][j]);        }        for(int i=0;i<n;i++) scanf("%lld",&arr[n-i-1]);        for(int i=0;i<n;i++) arr[i]--;        for(int mid=0;mid<n;mid++) {            int u=arr[mid];            ll temp=0;            for(int e=0;e<n;e++) {                for(int s=0;s<n;s++) {                    int a=arr[s];                    int b=arr[e];                    dp[a][b]=_min(dp[a][b],dp[a][u]+dp[u][b]);                    if(s<=mid && e<=mid) temp+=dp[a][b];                }            }            ans[n-mid-1]=temp;        }        for(int i=0;i<n;i++) printf("%I64d ",ans[i]);        printf("\n");    }    return 0;}