POJ 3686 The Windy's 最小费用最大流-白红宇

POJ 3686 The Windy's 最小费用最大流

阅读量：6879 次

发布时间：2019-06-26

本文共 2587 字，大约阅读时间需要 8 分钟。

每个工厂拆成N个工厂，费用分别为1~N倍原费用。

//#pragma comment(linker, "/STACK:1024000000,1024000000")#include
      
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                    using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair
                    
                      pii;#define pb(a) push(a)#define INF 0x1f1f1f1f#define lson idx<<1,l,mid#define rson idx<<1|1,mid+1,r#define PI 3.1415926535898template
                     
                       T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c));}template
                      
                        T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c));}void debug() {#ifdef ONLINE_JUDGE#else freopen("in.txt","r",stdin); // freopen("d:\\out1.txt","w",stdout);#endif}int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!='\n')return ch; } return EOF;}const int maxn = 55;int N, M;int cost[maxn][maxn];void Input(){ scanf("%d%d", &N, &M); for(int i = 1; i <= N; i++) for(int j = 1; j <= M; j++) scanf("%d", &cost[i][j]);}struct Edge{ int from, to, cost, cap;};const int maxv = maxn * maxn + maxn;vector
                       
                         g[maxv];vector
                        
                          edge;int n,s,t;void add(int from, int to, int cost, int cap){ edge.push_back((Edge){ from, to, cost, cap}); g[from].push_back(edge.size() - 1); edge.push_back((Edge){to, from, -cost, 0}); g[to].push_back(edge.size() - 1);}void init(){ for(int i = 1; i <= n; i++) g[i].clear(); edge.clear();}void construct(){ n = N + N * M + 2; s = n - 1; t = n; init(); for(int i = 1; i <= N; i++) add(s, i, 0, 1); for(int k = 1; k <= N; k++) { for(int j = 1; j <= M; j++) { int id = N + (k - 1) * M + j; add(id, t, 0, 1); for(int i = 1; i <= N; i++) add(i, id, k * cost[i][j], 1); } }}int d[maxv];int inq[maxv];int road[maxv];int SPFA(){ memset(d, INF, sizeof(d)); memset(inq, 0, sizeof(inq)); queue
                         
                           q; q.push(s); d[s] = 0; road[s] = -1; while(!q.empty()) { int u = q.front(); q.pop(); inq[u] = false; for(int i = 0; i < g[u].size(); i++) { Edge &e = edge[g[u][i]]; if(e.cap > 0 && d[u] + e.cost < d[e.to]) { d[e.to] = d[u] + e.cost; road[e.to] = g[u][i]; if(!inq[e.to]) { inq[e.to] = true; q.push(e.to); } } } } return d[t] != INF;}int MCMF(){ int cost = 0; while(SPFA()) { cost += d[t]; for(int e = road[t]; e != -1; e = road[edge[e].from]) { edge[e].cap -= 1; edge[e^1].cap += 1; } } return cost;}int main(){ debug(); int t; scanf("%d", &t); for(int ca = 1; ca <= t; ca++) { Input(); construct(); printf("%.6f\n", MCMF() * 1.0 / N); } return 0;}