Codeforces Round #179 (Div. 2)---D. Greg and Graph (offline +floyd)

Greg has a weighed directed graph, consisting of n vertices. In this graph any pair of distinct vertices have an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:

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

The first line contains an integer n (1?≤?n?≤?500)-the number of vertices in the graph.

Next n lines contain n integers each-the graph adjacency matrix:the j-th number in the i-th line AIJ (1?≤?aij?≤?105,?ai I?=?0) represents the weight of the edge that goes from vertex i to vertex J.

The next line contains n distinct integers:x1,?x2,?...,? xn (1?≤?xi?≤?n)-the vertices that Greg deletes.
Output

Print n integers-the i-th number equals the required sum before the i-th step.

%lld specifier to read or write 64-bit integers in C + +. It is preferred to use the CIN, cout streams of the%i64d specifier.
Sample Test (s)
Input

1
0
1

Output

0

Input

2
0 5
4 0
1 2

Output

9 0

Input

4
0 3 1 1
6 0 400 1
2 4 0 1
1 1 1 0
4 1 2 3

Output

17 23 404 0

First off-line, the operation is stored well, and then reverse processing, each time with this point to relax, and then to count the current shortest

/************************************************************************* > File name:cf-179-d.cpp > Aut Hor:alex > Mail: [email protected] > Created time:2015 April 27 Monday 14:01 20 seconds ******************************* *****************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace STD;Const DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> PLL;Static Const intN =550; LL Dp[n][n]; LL Ans[n];intOpe[n];BOOLUse[n];intMain () {intN while(~scanf("%d", &n)) {memset(Use,0,sizeof(use)); for(inti =1; I <= N; ++i) { for(intj =1; J <= N; ++J) {scanf("%i64d", &dp[i][j]); }        } for(inti =1; I <= N; ++i) {scanf("%d", &ope[i]); } for(intk = n; K >=1; --K) {intp = ope[k]; USE[P] =1; for(inti =1; I <= N; ++i) { for(intj =1; J <= N;                ++J) {Dp[i][j] = min (Dp[i][j], dp[i][p] + dp[p][j]); }} Ans[k] =0; for(inti =1; I <= N; ++i) {if(Use[i]) { for(intj =1; J <= N; ++J) {if(Use[j])                        {Ans[k] + = dp[i][j]; }                    }                }            }        }printf("%i64d", ans[1]); for(inti =2; I <= N; ++i) {printf("%i64d", Ans[i]); }printf("\ n"); }return 0;}

Codeforces Round #179 (Div. 2)---D. Greg and Graph (offline +floyd)