晨跑

題目連接

https://www.luogu.org/problemnew/show/P2153

題解

求最大不相交路徑數(shù),并在路徑數(shù)最大前提下,求總路程最短.

太裸了.

求不相交路徑數(shù):將除$1,n$兩點外的所有點拆分,中間連一條容量為$1$,費用為$0$的邊.然后所有的原邊$u\to v$視作從$u$的出點連向$v$的入點的一條費用為路程,容量為$1$的邊.

從$1\to n$跑最小費用最大流即是答案.

代碼

// luogu-judger-enable-o2 #include <iostream> #include <algorithm> #include <cstring> #include <queue> #define pr(x) std::cout << #x << ':' << x << std::endl #define rep(i,a,b) for(int i = a;i <= b;++i)const int inf = 0x3f3f3f3f; const int mm = 111111; const int maxn = 999; int node,src,dest,edge; int ver[mm],flow[mm],cst[mm],nxt[mm]; int head[maxn],work[maxn],dis[maxn],q[maxn]; int tot_cost; void prepare(int _node,int _src,int _dest) {node=_node,src=_src,dest=_dest;for(int i=0; i<node; ++i)head[i]=-1;edge=0;tot_cost = 0; } void add_edge(int u,int v,int c,int cost) {ver[edge]=v,flow[edge]=c,nxt[edge]=head[u],cst[edge]=cost,head[u]=edge++;ver[edge]=u,flow[edge]=0,nxt[edge]=head[v],cst[edge]=-cost,head[v]=edge++; } int ins[maxn]; int pre[maxn]; bool Dinic_spfa() {memset(ins,0,sizeof(ins));memset(dis,inf,sizeof(dis));memset(pre,-1,sizeof(pre));std::queue<int> Q;Q.push(src);dis[src] = 0,ins[src] = 1;pre[src] = -1;while(!Q.empty()){int u = Q.front();Q.pop();ins[u] = 0;for(int e = head[u];e != -1;e = nxt[e]){int v = ver[e];if(!flow[e]) continue;if(dis[v] > dis[u] + cst[e]){dis[v] = dis[u] + cst[e];pre[v] = e;if(!ins[v]) ins[v] = 1,Q.push(v);}}}return dis[dest] < inf; } int Dinic_flow() {int i,ret=0,delta=inf;while(Dinic_spfa()){for(int i=pre[dest];i != -1;i = pre[ver[i^1]])delta = std::min(delta,flow[i]);for(int i=pre[dest];i != -1;i = pre[ver[i^1]])flow[i] -= delta,flow[i^1] += delta;ret+=delta;tot_cost += dis[dest]*delta;}return ret; } int n,m; int main() {std::ios::sync_with_stdio(false);std::cin >> n >> m;prepare(2*n,0,2*n-1);for(int i = 1;i <= n;++i) {if(i == 1 || i == n)add_edge(i-1,i-1+n,inf,0);else add_edge(i-1,i-1+n,1,0);}for(int i = 1;i <= m;++i) {int a,b,c;std::cin >> a >> b >> c;add_edge(a-1+n,b-1,1,c);}int myflow = Dinic_flow();std::cout << myflow << " " << tot_cost << std::endl;return 0; }

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