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\n",left);
    return 0;
}