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最大流+二分答案
#include<cstdio> #include<cstring> #include<string> #include<cmath> #include<vector> #include<queue> #include<algorithm> using namespace std; const int maxn = 300 + 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; int K,C,M; int R[maxn][maxn]; int Se[maxn*maxn]; 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; } void floyd() { for(int k=1; k<=K+C; k++) for(int i=1; i<=K+C; i++) for(int j=1; j<=K+C; j++) if(R[k][j]!=999999999&&R[i][k]!=999999999) if(R[i][j]>R[i][k]+R[k][j]) R[i][j]=R[i][k]+R[k][j]; } void solve() { s=0; t=K+C+1; int To=0; for(int i=K+1; i<=K+C; i++) for(int j=1; j<=K; j++) if(R[i][j]!=999999999) { Se[To]=R[i][j]; To++; } sort(Se,Se+To); int Min=0,Max=To-1; int Mid=(Min+Max)/2; while(1) { int Lim=Se[Mid]; init(); for(int i=K+1; i<=K+C; i++) for(int j=1; j<=K; j++) if(R[i][j]!=999999999) if(R[i][j]<=Lim) AddEdge(j,i,INF); for(int i=1; i<=K; i++) AddEdge(s,i,M); for(int i=K+1; i<=K+C; i++) AddEdge(i,t,1); if(dinic(s,t)<C) { Min=Mid+1; Mid=(Min+Max)/2; } else { Max=Mid; Mid=(Min+Max)/2; } if(Min==Max) break; } printf("%d\n",Se[Mid]); } int main() { while(~scanf("%d%d%d",&K,&C,&M)) { for(int i=1; i<=K+C; i++) for(int j=1; j<=K+C; j++) { scanf("%d",&R[i][j]); if(R[i][j]==0) R[i][j]=999999999; } floyd(); solve(); } return 0; }
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原文地址:http://www.cnblogs.com/zufezzt/p/4825845.html