题目:给出一个图,从图中找出两条最短路,使得边不重复。
分析:既然是最短路,那么,两条路径上的所有节点的入边(s,x)、出边(x,e)必定是最优的,即
dis[x] = dis[s]+edge_dis,dis[e] = dis[x]+edge_dis。
dis表示点x到节点1的最短路的距离。
所以建图时,先求一边最短路,根据最短路上的前驱(可能有多个)与该节点构成新图,对于新图求一遍最大流判断最大流是否大于等于2即可。
/* 题目:给出一个图,从图中找出两条最短路,使得边不重复。 分析:先求一遍最短路,最短路上的点可以构成一个新图,对于新图求一遍最大流,如果存在大于1的最大流,
则有解。 */
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <utility>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; typedef long long ll;
typedef unsigned long long ull; #define debug puts("here")
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)
#define pb push_back
#define RD(n) scanf("%d",&n)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)
#define All(vec) vec.begin(),vec.end()
#define MP make_pair
#define PII pair<int,int>
#define PQ priority_queue
#define cmax(x,y) x = max(x,y)
#define cmin(x,y) x = min(x,y)
#define Clear(x) memset(x,0,sizeof(x))
/* #pragma comment(linker, "/STACK:1024000000,1024000000") int size = 256 << 20; // 256MB
char *p = (char*)malloc(size) + size;
__asm__("movl %0, %%esp\n" :: "r"(p) ); */ /******** program ********************/ const int MAXN = 405;
const int MAXM = 200005;
const int INF = 1e9; bool use[MAXN];
int po[MAXN],tol;
int gap[MAXN],dis[MAXN],arc[MAXN],pre[MAXN],cur[MAXN];
int n,m,vs,vt;
int a[MAXN],b[MAXN]; vector< int > vec[MAXN]; struct node{
int y,f,next;
}edge[MAXM]; void Add(int x,int y,int f){
edge[++tol].y = y;
edge[tol].f = f;
edge[tol].next = po[x];
po[x] = tol;
}
void add(int x,int y,int f){
Add(x,y,f);
Add(y,x,0);
} bool sap(){
memset(dis,0,sizeof(dis));
memset(gap,0,sizeof(gap));
gap[0] = vt;
rep1(i,vt)
arc[i] = po[i]; int ans = 0;
int aug = INF;
int x = vs;
bool ok = false; while(dis[vs]<vt){
ok = false;
cur[x] = aug;
for(int i=arc[x];i;i=edge[i].next){
int y = edge[i].y;
if(edge[i].f>0&&dis[y]+1==dis[x]){
ok = true;
pre[y] = arc[x] = i;
aug = min(aug,edge[i].f);
x = y;
if(x==vt){
ans += aug;
while(x!=vs){
edge[pre[x]].f -= aug;
edge[pre[x]^1].f += aug;
x = edge[pre[x]^1].y;
}
aug = INF;
}
break;
}
}
if(ok)
continue;
int MIN = vt-1;
for(int i=po[x];i;i=edge[i].next)
if(edge[i].f>0&&dis[edge[i].y]<MIN){
MIN = dis[edge[i].y];
arc[x] = i;
}
if(--gap[dis[x]]==0)
break;
dis[x] = ++ MIN;
++ gap[dis[x]];
if(x!=vs){
x = edge[pre[x]^1].y;
aug = cur[x];
}
}
//cout<<cnt<<endl;
return ans>=2;
} void spfa(){ rep1(i,n)
dis[i] = INF;
Clear(use);
dis[1] = 0; queue<int> q;
q.push(1); while(!q.empty()){
int x = q.front();
q.pop();
use[x] = false;
for(int i=po[x];i;i=edge[i].next){
int y = edge[i].y;
int t = dis[x]+edge[i].f;
if(dis[y]>t){
dis[y] = t;
vec[y].clear();
vec[y].pb(x);
if(!use[y]){
use[y] = true;
q.push(y);
}
}else if(dis[y]==t)
vec[y].pb(x);
}
} Clear(po);
tol = 1;
vs = 1;
vt = n; //puts("**********************");
rep1(x,n)
foreach(i,vec[x]){
add( vec[x][i],x,1 );
//cout<<vec[x][i]<<" "<<x<<endl;
}
//puts("**********************");
} void dfs(int x){
if(x==1)
printf("1");
else
printf(" %d",x);
if(x==vt)return;
for(int i=po[x];i;i=edge[i].next){
//cout<<edge[i].y<<" "<<edge[i].f<<endl;
if( edge[i].f==0 && (i&1)==0 ){
dfs(edge[i].y);
edge[i].f = 1;
return;
}
}
} int main(){ #ifndef ONLINE_JUDGE
freopen("sum.in","r",stdin);
//freopen("sum.out","w",stdout);
#endif cin>>n>>m;
int x,y,z; Clear(po);
tol = 0; while(m--){
RD3(x,y,z);
Add(x,y,z);
Add(y,x,z);
}
spfa();
bool ok = sap(); if(ok){
//puts("YES"); dfs(vs);
puts(""); dfs(vs);
puts(""); }else
puts("No solution"); return 0;
}