codeforces 416E. President's Path( graph dp)

http://codeforces.com/problemset/problem/416/E

题目：

Good old Berland has n cities and m roads. Each road connects a pair of distinct cities and is bidirectional. Between any pair of cities, there is at most one road. For each road, we know its length.

We also know that the President will soon ride along the Berland roads from citys to cityt. Naturally, he will choose one of the shortest paths froms tot, but nobody can say for sure which path he will choose.

The Minister for Transport is really afraid that the President might get upset by the state of the roads in the country. That is the reason he is planning to repair the roads in the possible President's path.

Making the budget for such an event is not an easy task. For all possible distinct pairss, t (s < t) find the number of roads that lie on at least one shortest path froms to t.

Input

The first line of the input contains integers n, m (2 ≤ n ≤ 500,0 ≤ m ≤ n·(n - 1) / 2) — the number of cities and roads, correspondingly. Thenm lines follow, containing the road descriptions, one description per line. Each description contains three integersx_i, y_i, l_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i, 1 ≤ l_i ≤ 10⁶), where x_i, y_i are the numbers of the cities connected by thei-th road andl_i is its length.

Output

Print the sequence of integersc₁₂, c₁₃, ..., c_1n, c₂₃, c₂₄, ..., c_2n, ..., c_{n - 1, n}, where c_st is the number of roads that can lie on the shortest path froms tot. Print the elements of sequencec in the described order. If the pair of citiess andt don't have a path between them, thenc_st = 0.

Sample test(s)

Input

5 61 2 12 3 13 4 14 1 12 4 24 5 4

Output

1 4 1 2 1 5 6 1 2 1 

用floyd动态思想解决此问题，附带一张图说明：

先对每两点floyd最短化，然后记录能到达终点的位置点作为中点，统计总的线路。取中点的地方细细品味。。。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int N=130000,M=505,INF=1<<29;int map[M][M],dp[M],ans[M][M];struct node{    int u,v,len;}edge[N];void init(){    for(int i=1;i<M;i++){       for(int j=1;j<M;j++){           if(i==j) map[i][j]=0;           else map[i][j]=INF;       }    }    memset(ans,0,sizeof(ans));}int main(){    //freopen("cin.txt","r",stdin);    int n,m;//1<<30=1073741824  1<<29=536870912    int x,y,p;    while(cin>>n>>m){         init();         for(int i=0;i<m;i++){            scanf("%d%d%d",&x,&y,&p);            edge[i].u=x;            edge[i].v=y;            edge[i].len=p;            if(p<map[x][y]){                map[x][y]=map[y][x]=p;            }         }         for(int k=1;k<=n;k++){//floyd 最短化             for(int i=1;i<=n;i++){                  for(int j=1;j<=n;j++){//涉及端点                      map[i][j]=min(map[i][j],map[i][k]+map[k][j]);                  }//1<<30能导致溢出             }          }          /*for(int i=1;i<=n;i++){                          for(int j=1;j<=n;j++) cout<<map[i][j]<<" ";  cout<<endl;        }*/          for(int i=1;i<=n;i++){// 起点               memset(dp,0,sizeof(dp));               for(int j=0;j<m;j++){//中点 (也包含本身两个端点的情况)                    if(map[edge[j].u][i]==edge[j].len+map[edge[j].v][i])                    dp[edge[j].u]++;                    if(map[edge[j].v][i]==edge[j].len+map[edge[j].u][i])                    dp[edge[j].v]++;               }               for(int j=i+1;j<=n;j++){//终点                    for(int k=1;k<=n;k++){//中点                        if(map[i][j]==map[i][k]+map[k][j])  //相等即能到达。                        ans[i][j]+=dp[k];                    }                    if(i==n-1&&j==n) printf("%d\n",ans[i][j]);                    else printf("%d ",ans[i][j]);               }         }     }     return 0;}