[BZOJ1061][NOI2008]志愿者招募（费用流神题单纯形裸题）

题目描述

传送门

题解

关于费用流的神建图我无言以对。
转自神犇的博客：https://www.byvoid.com/blog/noi-2008-employee/
关于单纯形。。。裸题一道。
大家都用了毕生的经历写关于单纯形模板的解释，而窝坚信一句话：
如果你过几天就忘了，那么你没有真正掌握。——by reflash

代码

费用流

#include<iostream>#include<cstring>#include<cstdio>#include<queue>using namespace std;const int max_n=1e3+5;const int max_m=1e4+5;const int max_N=max_n+3;const int max_M=max_N*2+max_m;const int max_e=max_M*2;const int INF=1e9;int n,m,N,x,y,z,maxflow,mincost;int need[max_n];int point[max_N],next[max_e],v[max_e],remain[max_e],c[max_e],tot;int dis[max_N],last[max_N];bool vis[max_N];queue <int> q;inline void addedge(int x,int y,int cap,int z){    ++tot; next[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap; c[tot]=z;    ++tot; next[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0; c[tot]=-z;}inline int addflow(int s,int t){    int ans=INF,now=t;    while (now!=s){        ans=min(ans,remain[last[now]]);        now=v[last[now]^1];    }    now=t;    while (now!=s){        remain[last[now]]-=ans;        remain[last[now]^1]+=ans;        now=v[last[now]^1];    }    return ans;}inline bool bfs(int s,int t){    memset(dis,0x7f,sizeof(dis));    memset(vis,0,sizeof(vis));    dis[s]=0;    vis[s]=true;    while (!q.empty()) q.pop();    q.push(s);    while (!q.empty()){        int now=q.front(); q.pop();        vis[now]=false;        for (int i=point[now];i!=-1;i=next[i])          if (dis[v[i]]>dis[now]+c[i]&&remain[i]){            dis[v[i]]=dis[now]+c[i];            last[v[i]]=i;            if (!vis[v[i]]){                vis[v[i]]=true;                q.push(v[i]);            }          }    }    if (dis[t]>INF) return false;    int flow=addflow(s,t);    maxflow+=flow;    mincost+=flow*dis[t];    return true;}inline void major(int s,int t){    maxflow=0; mincost=0;    while (bfs(s,t));}int main(){    tot=-1;    memset(point,-1,sizeof(point));    memset(next,-1,sizeof(next));    scanf("%d%d",&n,&m);    N=n+3;    for (int i=1;i<=n;++i){        scanf("%d",&need[i]);        x=need[i]-need[i-1];        if (x>=0) addedge(1,1+i,x,0);        else addedge(1+i,N,-x,0);    }    x=-need[n];    if (x>=0) addedge(1,n+2,x,0);    else addedge(n+2,N,-x,0);    for (int i=1;i<=m;++i){        scanf("%d%d%d",&x,&y,&z);        addedge(1+x,1+y+1,INF,z);    }    for (int i=1;i<=n;++i)      addedge(1+i+1,1+i,INF,0);    major(1,N);    printf("%d\n",mincost);}

单纯形

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>using namespace std;const int max_n=1005;const int max_m=10005;const double eps=1e-7;const double inf=1e10;int n,m,L,R;double A[max_m][max_n],B[max_m],C[max_n],ans;inline void pivot(int l,int e){    B[l]/=A[l][e];    for (int i=1;i<=n;++i)      if (i!=e)        A[l][i]/=A[l][e];    A[l][e]=1/A[l][e];    for (int i=1;i<=m;++i)      if (i!=l&&fabs(A[i][e])>eps){        B[i]-=B[l]*A[i][e];        for (int j=1;j<=n;++j)          if (j!=e)            A[i][j]-=A[l][j]*A[i][e];        A[i][e]=-A[l][e]*A[i][e];      }    ans+=B[l]*C[e];    for (int i=1;i<=n;++i)      if (i!=e)        C[i]-=C[e]*A[l][i];    C[e]=-C[e]*A[l][e];}inline void simplex(){    int e;    while (1){        for (e=1;e<=n;++e)          if (C[e]>eps) break;        if (e==n+1) break;        double data=inf,t; int l;        for (int i=1;i<=m;++i)          if (A[i][e]>eps&&(t=B[i]/A[i][e])<data){            data=t;            l=i;          }        pivot(l,e);    }}int main(){    scanf("%d%d",&n,&m);    for (int i=1;i<=n;++i)      scanf("%lf",&C[i]);    for (int i=1;i<=m;++i){        scanf("%d%d%lf",&L,&R,&B[i]);        for (int j=L;j<=R;++j)          A[i][j]++;    }    simplex();    printf("%0.lf",ans);}

总结

ORZ hxy神犇——该人曾用费用流艹掉了学长的互测题（标解是单纯形，你们可以根据这道题的建图来脑补一下难度）