nwerc2013 A-Absurdistan Roads (最小生成树(kruskal)+最短路(floyd))
Absurdistan Roads
Time Limit: 5678/3456MS (Java/Others) Memory Limit: 65432/65432KB (Java/Others)
The people of Absurdistan discovered how to build roads only last year. After the discovery, every city decided to build their own road connecting
their city with another city. Each newly built road can be used in both directions.
Absurdistan is full of surprising coincidences. It took all
possible to travel from every city to every other city using the newly built roads.
You bought a tourist guide which does not have a map of the country with the new roads. It only contains a huge table with the shortest distances
between all pairs of cities using the newly built roads. You would like to know between which pairs of cities there are roads and how long they are,
because you want to reconstruct the map of the
You get a table of shortest distances between all pairs of cities in Absurdistan using the
reconstruct the road network of Absurdistan. There might be multiple road networks with
but you are happy with any one of those networks.
Input
For each test case:
- A line containing an integer
N (2≤N≤2000) -- the number of cities and roads. N lines withN numbers each. Thej -th number of thei -th line is the shortest distance from cityi to cityj . All distances between two- distinct cities will be positive and at most
1000000 . - The distance from
i toi will always be0 and the distance fromi toj will be the same as the distance fromj toi .
Output
For each test case:
- Print
N lines with three integers 'a b c ' denoting that there is a road between cities1≤a≤N and1≤b≤N of length1≤c≤1000000 , - where
a≠b . If there are multiple solutions, you can print any one and you can print the roads in any order. At least one solution is - guaranteed to exist.
Print a blank line between every two test cases.
Sample input and output
Sample Input | Sample Output |
---|---|
4 0 1 2 1 1 0 2 1 2 2 0 1 1 1 1 0 4 0 1 1 1 1 0 2 2 1 2 0 2 1 2 2 0 3 0 4 1 4 0 3 1 3 0 |
2 1 1 4 1 1 4 2 1 4 3 1 2 1 1 3 1 1 4 1 1 2 1 1 3 1 1 2 1 4 3 2 3 |
Source
题意:给你一个邻接矩阵代表最短路的情况,让你给出n条原始的边,使得这个最短路成立。
方法:要找的都是必要的边。先构造1棵最小生成树,这n-1条边是部分答案,然后是找最后一条边。用floyd求出任意两点间的距离,
找出未满足最短距离并且这两个点的最短距离最小的边,如果都已满足最短距离则随意添加一条重边即可。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #pragma comment (linker,"/STACK:102400000,102400000") #define maxn 1005 #define MAXN 2005 #define mod 1000000009 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-6 typedef long long ll; using namespace std; struct Edge { int u,v,len; }edge[4000005]; int dist[2005],mp[2005][2005]; int father[2005]; int N,num; int cmp(Edge a,Edge b) { return a.len<b.len; } void init(int n) { memset(mp,INF,sizeof(mp)); for (int i=1;i<=n;i++) father[i]=i; } int find_father(int x) { if (x!=father[x]) father[x]=find_father(father[x]); return father[x]; } void Kruskal() { int i,j,sum=0; init(N); for (i=0;i<num;i++) { int fa=find_father(edge[i].u); int fb=find_father(edge[i].v); if (fa!=fb) { father[fa]=fb; mp[ edge[i].u ][ edge[i].v ]=mp[ edge[i].v ][ edge[i].u ]=edge[i].len; printf("%d %d %d\n",edge[i].u,edge[i].v,edge[i].len); sum++; if (sum>=N-1) break; } } } void floyd() { int i,j,k; for (k=1;k<=N;k++) for (i=1;i<=N;i++) for (j=1;j<=N;j++) { if (mp[i][k]==INF) break; mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]); } } int main() { int i,j,flag=0; while (~scanf("%d",&N)) { num=0; int w; if (flag) printf("\n"); flag=1; for (i=1;i<=N;i++) for (j=1;j<=N;j++) { scanf("%d",&w); if (i<=j) continue; edge[num].u=i; edge[num].v=j; edge[num++].len=w; } sort(edge,edge+num,cmp); Kruskal(); floyd(); for (i=0;i<num;i++) { if (edge[i].len!=mp[ edge[i].u ][ edge[i].v ]) { printf("%d %d %d\n",edge[i].u,edge[i].v,edge[i].len); break; } } if (i==num) printf("%d %d %d\n",edge[0].u,edge[0].v,edge[0].len); } return 0; } /* 4 0 1 2 1 1 0 2 1 2 2 0 1 1 1 1 0 4 0 1 1 1 1 0 2 2 1 2 0 2 1 2 2 0 3 0 4 1 4 0 3 1 3 0 */