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 " :: "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; }