(权限题)NFLSoj #113: 或负 题解

一眼网络流
考虑超级源点向第i个长者连流量为L[i]的边，每个长者拆成D天，向每一天连N-sum[i][j]，sum[i][j]表示他当天必须用来准备批判会的时间
每个长者的每一天专门再拆一个点管理长者的午休，连Hl-Hr-sum1[i][j]，sum1[i][j]表示他当天午休时间内必须用来准备批判会的时间
最后对应的点向批判会连边

*数组不要开小了！！！

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <utility>
#include <cctype>
#include <algorithm>
#include <bitset>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <cmath>
#define LL long long
#define LB long double
#define x first
#define y second
#define Pair pair<int,int>
#define pb push_back
#define pf push_front
#define mp make_pair
#define LOWBIT(x) x & (-x)
using namespace std;

const int MOD=1e9+7;
const LL LINF=2e16;
const int INF=2e9;
const int magic=348;
const double eps=1e-10;
const double pi=acos(-1);

inline int getint()
{
    char ch;int res;bool f;
    while (!isdigit(ch=getchar()) && ch!='-') {}
    if (ch=='-') f=false,res=0; else f=true,res=ch-'0';
    while (isdigit(ch=getchar())) res=res*10+ch-'0';
    return f?res:-res;
}

int Z,D,H,N,Hl,Hr;
int L[148],R[148][148],F[148][148][148],sum[148][148],sum1[148][148];
int t;

int head[200048],to[800048],nxt[800048],f[800048],tot=1;
inline void addedge(int s,int t,int cap)
{
    to[++tot]=t;nxt[tot]=head[s];head[s]=tot;f[tot]=cap;
    to[++tot]=s;nxt[tot]=head[t];head[t]=tot;f[tot]=0;
}

int depth[200048];queue<int> q;
inline bool bfs()
{
    int i,x,y;
    for (i=0;i<=t;i++) depth[i]=-1;
    depth[0]=0;q.push(0);
    while (!q.empty())
    {
        x=q.front();q.pop();
        for (i=head[x];i;i=nxt[i])
        {
            y=to[i];
            if (depth[y]==-1 && f[i])
            {
                depth[y]=depth[x]+1;
                q.push(y);
            }
        }
    }
    if (depth[t]==-1) return false; else return true;
}

inline int dfs(int x,int maxf)
{
    if (x==t || !maxf) return maxf;
    int i,y,now,minf,ans=0;
    for (i=head[x];i;i=nxt[i])
    {
        y=to[i];
        if (depth[y]==depth[x]+1 && f[i])
        {
            minf=min(maxf-ans,f[i]);
            now=dfs(y,minf);
            f[i]-=now;f[i^1]+=now;ans+=now;
        }
        if (ans>=maxf) return ans;
    }
    if (!ans) depth[x]=0;
    return ans;
}

int main ()
{
    int i,j,k;
    Z=getint();D=getint();H=getint();N=getint();Hl=getint();Hr=getint();
    for (i=1;i<=Z;i++) L[i]=getint();
    for (i=1;i<=D;i++)
        for (j=1;j<=H;j++)
            R[i][j]=getint();
    for (i=1;i<=Z;i++)
        for (j=1;j<=D;j++)
        {
            sum[i][j]=N;
            for (k=1;k<=H;k++)
            {
                F[i][j][k]=getint();
                sum[i][j]-=(F[i][j][k]==0);
                if (Hl<=k && k<=Hr) sum1[i][j]+=(F[i][j][k]==1);
            }
            if (sum1[i][j]==0) {printf("No\n");return 0;}
            if (sum[i][j]<0) {printf("No\n");return 0;}
        }
    t=Z+D*Z+D*Z+D*H+1;
    for (i=1;i<=Z;i++) addedge(0,i,L[i]);
    for (i=1;i<=Z;i++)
        for (j=1;j<=D;j++)
            addedge(i,Z+(i-1)*D+j,sum[i][j]);
    for (i=1;i<=Z;i++)
        for (j=1;j<=D;j++)
            addedge(Z+(i-1)*D+j,Z+D*Z+(i-1)*D+j,sum1[i][j]-1);
    for (i=1;i<=Z;i++)
        for (j=1;j<=D;j++)
            for (k=1;k<=H;k++)
            {
                if (!F[i][j][k]) continue;
                if (Hl<=k && k<=Hr)
                    addedge(Z+D*Z+(i-1)*D+j,Z+D*Z+D*Z+(j-1)*H+k,1);
                else
                    addedge(Z+(i-1)*D+j,Z+D*Z+D*Z+(j-1)*H+k,1);
            }
    for (i=1;i<=D;i++)
        for (j=1;j<=H;j++)
            addedge(Z+D*Z+D*Z+(i-1)*H+j,t,R[i][j]);
    int ans=0,need=0;
    for (i=1;i<=D;i++)
        for (j=1;j<=H;j++)
            need+=R[i][j];
    while (bfs()) ans+=dfs(0,2e9);
    if (ans==need) printf("Yes\n"); else printf("No\n");
    return 0;
}