BZOJ 3993 Luogu P3324 [SDOI2015]星际战争 (最大流、二分答案)

字符串终于告一段落了！

题目链接: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=3993

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

网络流从最水的开始做。。。

题解: 二分答案ans, 然后可以得到每个攻击者在ans时间内最多产生的总伤害，从起点往攻击者连边容量为此值，从每个攻击者往能攻击的防御者连边容量为(+inf), 从每个防御者往终点连边边权为其装甲值，求最大流，判断其是否等于装甲值之和即可。

时间复杂度(O(MaxFlow(n,n^2)))

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

const int N = 102;
const int M = 2600;
const double INF = 1e7;
const double EPS = 1e-8;

namespace MaxFlow
{
	struct Edge
	{
		int v,nxt,rev; double w;
	} e[(M<<1)+3];
	int fe[N+3];
	int te[N+3];
	int dep[N+3];
	int que[N+3];
	int n,en;
	
	void clear()
	{
		for(int i=1; i<=en; i++)
		{
			e[i].v = e[i].nxt = e[i].rev = 0; e[i].w = 0.0;
		}
		for(int i=1; i<=n; i++) fe[i] = 0;
		n = en = 0;
	}
	void addedge(int u,int v,double w)
	{
		en++; e[en].v = v; e[en].w = w;
		e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;
		en++; e[en].v = u; e[en].w = 0;
		e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;
	}
	bool bfs()
	{
		for(int i=1; i<=n; i++) dep[i] = 0;
		int head = 1,tail = 1; que[tail] = 1; dep[1] = 1;
		while(head<=tail)
		{
			int u = que[head]; head++;
			for(int i=fe[u]; i; i=e[i].nxt)
			{
				if(dep[e[i].v]==0 && e[i].w>EPS)
				{
					dep[e[i].v] = dep[u]+1;
					tail++; que[tail] = e[i].v;
				}
			}
		}
		return dep[2]!=0;
	}
	double dfs(int u,double cur)
	{
		if(u==2) return cur;
		double rst = cur;
		for(int i=te[u]; i; i=e[i].nxt)
		{
			if(dep[e[i].v]==dep[u]+1 && e[i].w>0 && rst>0)
			{
				double flow = dfs(e[i].v,min(rst,e[i].w));
				if(flow>EPS)
				{
					rst -= flow; e[i].w -= flow; e[e[i].rev].w += flow;
					if(e[i].w>EPS) te[u] = i;
					if(rst==0) return cur;
				}
			}
		}
		if(abs(cur-rst)<EPS) dep[u] = 0;
		return cur-rst;
	}
	double dinic(int _n)
	{
		n = _n;
		double ret = 0.0;
		while(bfs())
		{
			for(int i=1; i<=n; i++) te[i] = fe[i];
			ret += dfs(1,INF);
		}
		return ret;
	}
}

int a[N+3],b[N+3];
int c[N+3][N+3];
int n,m;

int main()
{
	scanf("%d%d",&n,&m); double std = 0.0;
	for(int i=1; i<=n; i++) {scanf("%d",&a[i]); std += (double)a[i];}
	for(int i=1; i<=m; i++) scanf("%d",&b[i]);
	for(int i=1; i<=m; i++)
	{
		for(int j=1; j<=n; j++) scanf("%d",&c[i][j]);
	}
	double left = 0.0,right = INF;
	while(right-left>1e-5)
	{
		double mid = (left+right)*0.5;
		MaxFlow::clear();
		for(int i=1; i<=m; i++)
		{
			MaxFlow::addedge(1,i+2,mid*(double)b[i]);
		}
		for(int i=1; i<=n; i++)
		{
			MaxFlow::addedge(i+m+2,2,(double)a[i]);
		}
		for(int i=1; i<=m; i++)
		{
			for(int j=1; j<=n; j++)
			{
				if(c[i][j]==1)
				{
					MaxFlow::addedge(i+2,j+m+2,INF);
				}
			}
		}
		double ans = MaxFlow::dinic(m+n+2);
		if(abs(ans-std)<EPS)
		{
			right = mid;
		}
		else
		{
			left = mid;
		}
	}
	printf("%.6lf
",left);
	return 0;
}