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HDU 5889 Barricade
最短路,最小割,网络流。
可以根据$dis[u]+1$与$dis[v]$的大小关系判断$<u,v>$是否为最短路上的边,可以处理出一个只包含最短路的$DAG$,然后求这个$DAG$的最小割就可以了。
#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<iostream>using namespace std;typedef long long LL;const double pi=acos(-1.0),eps=1e-6;void File(){ freopen("D:\\in.txt","r",stdin); freopen("D:\\out.txt","w",stdout);}template <class T>inline void read(T &x){ char c=getchar(); x=0; while(!isdigit(c)) c=getchar(); while(isdigit(c)) {x=x*10+c-‘0‘; c=getchar();}}const int maxn = 30000 + 10;const int INF = 0x7FFFFFFF;struct Edge{ int from, to, cap, flow; Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f){}};vector<Edge>edges;vector<int>G[maxn];bool vis[maxn];int d[maxn];int cur[maxn];int n, m, s, t;void init(){ for (int i = 0; i < maxn; i++) G[i].clear(); edges.clear();}void AddEdge(int from, int to, int cap){ edges.push_back(Edge(from, to, cap, 0)); edges.push_back(Edge(to, from, 0, 0)); int w = edges.size(); G[from].push_back(w - 2); G[to].push_back(w - 1);}bool BFS(){ memset(vis, 0, sizeof(vis)); queue<int>Q; Q.push(s); d[s] = 0; vis[s] = 1; while (!Q.empty()) { int x = Q.front(); Q.pop(); for (int i = 0; i<G[x].size(); i++) { Edge e = edges[G[x][i]]; if (!vis[e.to] && e.cap>e.flow) { vis[e.to] = 1; d[e.to] = d[x] + 1; Q.push(e.to); } } } return vis[t];}int DFS(int x, int a){ if (x == t || a == 0) return a; int flow = 0, f; for (int &i = cur[x]; i<G[x].size(); i++) { Edge e = edges[G[x][i]]; if (d[x]+1 == d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0) { edges[G[x][i]].flow+=f; edges[G[x][i] ^ 1].flow-=f; flow+=f; a-=f; if(a==0) break; } } if(!flow) d[x] = -1; return flow;}int dinic(int s, int t){ int flow = 0; while (BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow;}int h[maxn],sz,T;struct X{ int u,v,w,nx;}ee[maxn];int dis[maxn],flag[maxn];void add(int a,int b,int c){ ee[sz].u=a; ee[sz].v=b; ee[sz].w=c; ee[sz].nx=h[a]; h[a]=sz++; }void spfa(){ for(int i=1;i<=n;i++) dis[i]=INF ,flag[i]=0; dis[1]=0; queue<int>Q; Q.push(1); flag[1]=1; while(!Q.empty()) { int top=Q.front(); Q.pop(); flag[top]=0; for(int i=h[top];i!=-1;i=ee[i].nx) { if(dis[top]+1<dis[ee[i].v]) { dis[ee[i].v]=dis[top]+1; if(flag[ee[i].v]==0) { flag[ee[i].v]=1; Q.push(ee[i].v); } } } }}int main(){ scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); sz=0; memset(h,-1,sizeof h); for(int i=1;i<=m;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); add(u,v,w); add(v,u,w); } spfa(); init(); for(int i=0;i<sz;i++) { if(dis[ee[i].u]+1==dis[ee[i].v]) { AddEdge(ee[i].u,ee[i].v,ee[i].w); } } s=1; t=n; printf("%d\n",dinic(s,t)); } return 0;}
HDU 5889 Barricade
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