【SDOI2015】星际战争

题面

https://www.luogu.org/problemnew/show/P3324

题解

水题。

二分答案+最大流判可行。

• 注意最大流解决伤血模型的运用。

// luogu-judger-enable-o2
#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
#define ri register int
#define N 505
#define INF 1000000007
#define T (n+m+1)
#define eps 1e-5

using namespace std;

int n,m;
vector<int> to,ed[N];
vector<double> w;
int cur[N],d[N];
double sum;
int g[N][N],a[N],b[N];

inline int read() {
  int f=0,ret=0; char ch=getchar();
    while (ch<'0' || ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') ret*=10,ret+=(ch-'0'),ch=getchar();
    return f?-ret:ret;
}

void add_edge(int u,int v,double w1,double w2) {
  to.push_back(v); w.push_back(w1); ed[u].push_back(to.size()-1);
  to.push_back(u); w.push_back(w2); ed[v].push_back(to.size()-1);
}

bool bfs() {
  queue<int> q;
  memset(d,0x3f,sizeof(d));
  d[0]=0; q.push(0);
  while (!q.empty()) {
    int x=q.front(); q.pop();
    for (ri i=0,l=ed[x].size();i<l;i++) {
      int e=ed[x][i];
      if (w[e]>eps && d[x]+1<d[to[e]]) {
        d[to[e]]=d[x]+1;
        q.push(to[e]);
      }
    }
  }
  return d[T]<=1000000;
}

double dfs(int x,double limit) {
  if (x==T || !limit) return limit;
  double tot=0;
  for (ri &i=cur[x];i<ed[x].size();i++) {
    int e=ed[x][i];
    if (d[to[e]]==d[x]+1 && w[e]>eps) {
      double f=dfs(to[e],min(limit,w[e]));
      if (f<eps) continue;
      w[e]-=f; w[1^e]+=f; 
      tot+=f; limit-=f;
      if (limit<eps) return tot;
    }
  }
  return tot;
}

double dinic() {
  double ret=0;
  while (bfs()) {
    memset(cur,0,sizeof(cur));
    ret+=dfs(0,INF);
  }
  return ret;
}

bool check(double t) {
  w.clear(); to.clear();
  for (ri i=0;i<=n+m+1;i++) ed[i].clear();
  for (ri i=1;i<=m;i++) add_edge(0,i,t*b[i],0);
  for (ri i=1;i<=n;i++) add_edge(m+i,T,a[i],0);
  for (ri i=1;i<=m;i++)
     for (ri j=1;j<=n;j++) if (g[i][j]) add_edge(i,m+j,INF,0);
  if (dinic()+eps>sum) return 1; else return 0;
}

int main() {
  n=read(); m=read();
  for (ri i=1;i<=n;i++) a[i]=read();
  sum=0.0;
  for (ri i=1;i<=n;i++) sum+=a[i];
  for (ri i=1;i<=m;i++) b[i]=read();
  for (ri i=1;i<=m;i++)
     for (ri j=1;j<=n;j++) g[i][j]=read();
  double lb=0,rb=10007; double ans=-1;
  while (rb-lb>1e-4) {
    double mid=(lb+rb)/2;
     if (check(mid)) ans=mid,rb=mid; else lb=mid;
  }
  printf("%.4lf",ans);
  return 0;
}