HDU 4725 The Shortest Path in Nya Graph(最短路拆点)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4725
题意:n个点,某个点属于某一层。共有n层。第i层的点到第i+1层的点和到第i-1层的点的代价均是C。另外有一类边连接两个点u、v,代价w。求1到n的最短路。
思路:拆点。n个点不动,编号1到n。将n层每层拆成两个点,第i层拆成n+i*2-1,n+i*2。相邻的层连边(n+i*2-1,n+(i+1)*2,C),(n+(i+1)*2-1,n+i*2,C)。若顶点u属于第i层,连边(u,n+i*2-1,0),(n+i*2,u,0)。再加上另外的那种边。最后用优先队列跑最短路。
vector<pair<int,int> > g[N]; int a[N]; int f[N],n,m,C; struct node { int u,cost; node(){} node(int _u,int _cost) { u=_u; cost=_cost; } int operator<(const node &a) const { return cost>a.cost; } }; int cal() { priority_queue<node> Q; Q.push(node(1,0)); int i; FOR1(i,n*3) f[i]=INF; f[1]=0; while(!Q.empty()) { int u=Q.top().u; Q.pop(); if(u==n) return f[u]; FOR0(i,SZ(g[u])) { int v=g[u][i].first; int w=g[u][i].second; if(f[u]+w<f[v]) { f[v]=f[u]+w; Q.push(node(v,f[v])); } } } return -1; } void add(int u,int v,int w) { g[u].pb(MP(v,w)); } int main() { int num=0; rush() { RD(n,m,C); int i; FOR1(i,n*3) g[i].clear(); FOR1(i,n-1) { add(n+i*2-1,n+(i+1)*2,C); add(n+(i+1)*2-1,n+i*2,C); } FOR1(i,n) { RD(a[i]); add(i,n+a[i]*2-1,0); add(n+a[i]*2,i,0); } FOR1(i,m) { int u,v,w; RD(u,v,w); add(u,v,w); add(v,u,w); } int ans=cal(); printf("Case #%d: %d\n",++num,ans); } }
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