【SCOI2007】【最小费用最大流】修车

需要一定思考才能建出模型的费用流。

设有n辆车，m个维修人员

每辆车看作一个点u，连边s -> u，容量为1，费用为0

然后将每个维修人员拆成n个点，连边u-> v，容量为1，费用为k * time[u][v](k表示拆分的n个点中的第k个点)，实际意义为第v个维修人员在倒数第k个位置修第u辆车

最后对于所有与维修人员相关的点，连边v -> t，容量为1，费用为0

做一次最小费用最大流就是最小等待时间了，题目要求平均最小时间，再除以总人数n就行了

#include<cstdio>#include<cstring>using namespace std;const int inf = 0x3f3f3f3f;const int maxp = 100;const int maxn = 1000;const int maxm = 200000;struct Edge{int pos,c,d,w;int next;}E[maxm];int head[maxn],time[maxp][maxp];int dis[maxn],pre[maxn];int que[maxm];bool vis[maxn];int s,t,NE;int n,m;void init(){freopen("bzoj1070.in","r",stdin);freopen("bzoj1070.out","w",stdout);}void add(int u,int v,int c,int w){E[NE].pos = v;E[NE].d = u;E[NE].c = c;E[NE].w = w;E[NE].next = head[u];head[u] = NE++;E[NE].pos = u;E[NE].d = v;E[NE].c = 0;E[NE].w = -w;E[NE].next = head[v];head[v] = NE++;}void readdata(){scanf("%d%d",&m,&n);for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){scanf("%d",&time[i][j]);}}}void build_map(){memset(head,-1,sizeof(head));memset(E,0,sizeof(E));s = 0,t = n * (m + 1) + 1;NE = 0;for(int i = 1;i <= n;i++)add(s,i,1,0);for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++)for(int k = 1;k <= n;k++)add(i,j * n + k,1,k * time[i][j]);}for(int i = n + 1;i < t;i++)add(i,t,1,0);}bool spfa(){memset(dis,0x3f,sizeof(dis));memset(vis,false,sizeof(vis));memset(pre,-1,sizeof(pre));int l = 0,r = 0; dis[s] = 0;que[r++] = s;vis[s] = true;while(l < r){int u = que[l++];vis[u] = false;for(int i = head[u];i != -1;i = E[i].next){int v = E[i].pos;if(E[i].c && dis[u] + E[i].w < dis[v]){pre[v] = i;dis[v] = dis[u] + E[i].w;if(!vis[v]){que[r++] = v;vis[v] = true;}}}}if(dis[t] == inf)return false;else return true;}int MCMF(){int ret = 0;while(spfa()){int min = inf;int u = t;while(u != s){if(E[pre[u]].c < min)min = E[pre[u]].c;u = E[pre[u]].d;}u = t;while(u != s){E[pre[u]].c -= min;E[pre[u]^1].c += min;u = E[pre[u]].d;}ret += min * dis[t];}return ret;}void solve(){build_map();double ans = (double)MCMF() / (double)n;printf("%.2lf\n",ans);}int main(){init();readdata();solve();return 0;}