CodeForces 25CRoads in Berland【floyd过】

Roads in Berland

Description 
There are n cities numbered from 1 to n in Berland. Some of them are connected by two-way roads. Each road has its own length — an integer number from 1 to 1000. It is known that from each city it is possible to get to any other city by existing roads. Also for each pair of cities it is known the shortest distance between them. Berland Government plans to build k new roads. For each of the planned road it is known its length, and what cities it will connect. To control the correctness of the construction of new roads, after the opening of another road Berland government wants to check the sum of the shortest distances between all pairs of cities. Help them — for a given matrix of shortest distances on the old roads and plans of all new roads, find out how the sum of the shortest distances between all pairs of cities changes after construction of each road.

Input 
The first line contains integer n (2 ≤ n ≤ 300) — amount of cities in Berland. Then there follow n lines with n integer numbers each — the matrix of shortest distances. j-th integer in the i-th row — di, j, the shortest distance between cities i and j. It is guaranteed that di, i = 0, di, j = dj, i, and a given matrix is a matrix of shortest distances for some set of two-way roads with integer lengths from 1 to 1000, such that from each city it is possible to get to any other city using these roads.

Next line contains integer k (1 ≤ k ≤ 300) — amount of planned roads. Following k lines contain the description of the planned roads. Each road is described by three space-separated integers ai, bi, ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 1000) — ai and bi — pair of cities, which the road connects, ci — the length of the road. It can be several roads between a pair of cities, but no road connects the city with itself.

Output 
Output k space-separated integers qi (1 ≤ i ≤ k). qi should be equal to the sum of shortest distances between all pairs of cities after the construction of roads with indexes from 1 to i. Roads are numbered from 1 in the input order. Each pair of cities should be taken into account in the sum exactly once, i. e. we count unordered pairs.

Sample Input 
Input 
2 
0 5 
5 0 
1 
1 2 3 
Output 
3 
Input 
3 
0 4 5 
4 0 9 
5 9 0 
2 
2 3 8 
1 2 1 
Output 
17 12

题目大意：每一次改变一个边的长度，输出各个点之间最短距离和、

一看各个点之间的最短路径和，floyd明显符合条件、

floyd递推最短路、但是300^4明显会超时，所以我们这里要优化代码，只判断和u，v（改变边长的两个点）这两个特殊的点就行了、

如果从j到k，能够通过u这个点把路径变短，那么就出现了一次优化路径。在经过u，v这两个点的判断的时候，不免就会优化到u，v这条边，所以代码这样写就ok了：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;#define ll long long intll map[305][305];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                scanf("%I64d",&map[i][j]);            }        }        int k;        scanf("%d",&k);        while(k--)        {            int u,v;            ll val;            scanf("%d%d%I64d",&u,&v,&val);            if(val<map[u][v])            {                map[u][v]=val;                map[v][u]=val;                for(int j=1;j<=n;j++)                {                    for(int k=1;k<=n;k++)                    {                        map[k][j]=map[j][k]=min(map[j][k],map[j][u]+map[u][k]);                    }                }                for(int j=1;j<=n;j++)                {                    for(int k=1;k<=n;k++)                    {                        map[k][j]=map[j][k]=min(map[j][k],map[j][v]+map[v][k]);                    }                }            }            ll sum=0;            for(int i=1;i<=n;i++)            {                for(int j=1;j<=n;j++)                {                    sum+=map[i][j];                }            }            printf("%I64d\n",sum/2);        }    }}