Shoot the Bullet(ZOJ3229)（有源汇上下界最大流）

描述

ensokyo is a world which exists quietly beside ours, separated by a mystical border. It is a utopia where humans and other beings such as fairies, youkai(phantoms), and gods live peacefully together. Shameimaru Aya is a crow tengu with the ability to manipulate wind who has been in Gensokyo for over 1000 years. She runs the Bunbunmaru News - a newspaper chock-full of rumors, and owns the Bunkachou - her record of interesting observations for Bunbunmaru News articles and pictures of beautiful danmaku(barrange) or cute girls living in Gensokyo. She is the biggest connoisseur of rumors about the girls of Gensokyo among the tengu. Her intelligence gathering abilities are the best in Gensokyo!

During the coming n days, Aya is planning to take many photos of m cute girls living in Gensokyo to write Bunbunmaru News daily and record at least G_x photos of girl x in total in the Bunkachou. At the k-th day, there are C_k targets, T_k1, T_k2, ..., T_{kC_k}. The number of photos of target T_ki that Aya takes should be in range [L_ki, R_ki], if less, Aya cannot write an interesting article, if more, the girl will become angry and use her last spell card to attack Aya. What's more, Aya cannot take more than D_k photos at the k-th day. Under these constraints, the more photos, the better.

Aya is not good at solving this complex problem. So she comes to you, an earthling, for help.

Input

There are about 40 cases. Process to the end of file.

Each case begins with two integers 1 <= n <= 365, 1 <= m <= 1000. Then m integers, G₁, G₂, ..., G_m in range [0, 10000]. Then n days. Each day begins with two integer 1 <= C <= 100, 0 <= D <= 30000. Then C different targets. Each target is described by three integers, 0 <= T< m, 0 <= L <= R <= 100.

Output

For each case, first output the number of photos Aya can take, -1 if it's impossible to satisfy her needing. If there is a best strategy, output the number of photos of each girl Aya should take at each day on separate lines. The output must be in the same order as the input. If there are more than one best strategy, any one will be OK.

Output a blank line after each case.

Sample Input

Sample Output

题意：

大致是说一个男生要个其他女生拍照，总共拍N天，总共有M个女生，

接下来一行是说M个女生,每个女生至少要拍照的数量

然后每天都有自己的拍照计划，给多少女生拍照，今天拍照的总数量

接下来一行是对于女生x,这一天拍照的上限L和下限R

题解：

有源汇上下界最大流，先建图，首先源点和每天连线，上届是这一天拍照总数，下界是0，每一天再和当天拍照的女生建边，上下界是L,R，每个女生和汇点建边，上下界是INf,和给这个女生拍照的最少数量。

然后转换为无源汇上下界最大流，建立超级源点和汇点，跑最大流看是否满流，是则，再跑一遍，求出最大流，否输出-1

#pragma GCC optimize(2)

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <algorithm>

#define inf 0x3f3f3f3f

#define ll long long

#define MAXN 30000

using namespace std;

int n,m;//点数、边数

int sp,tp;//原点、汇点

struct node

{

    int v,next;

    ll cap;

}mp[MAXN*10];

int pre[MAXN],dis[MAXN],cur[MAXN];//cur为当前弧优化，dis存储分层图中每个点的层数（即到原点的最短距离），pre建邻接表

int cnt=0;

void init()//不要忘记初始化

{

    cnt=0;

    memset(pre,-1,sizeof(pre));

}

void add(int u,int v,int w)//加边

{

    mp[cnt].v=v;

    mp[cnt].cap=w;

    mp[cnt].next=pre[u];

    pre[u]=cnt++;

    mp[cnt].v=u;

    mp[cnt].cap=0;

    mp[cnt].next=pre[v];

    pre[v]=cnt++;

}

bool bfs()//建分层图

{

    memset(dis,-1,sizeof(dis));

    queue<int>q;

    while(!q.empty())

        q.pop();

    q.push(sp);

    dis[sp]=0;

    int u,v;

    while(!q.empty())

    {

        u=q.front();

        q.pop();

        for(int i=pre[u];i!=-1;i=mp[i].next)

        {

            v=mp[i].v;

            if(dis[v]==-1&&mp[i].cap>0)

            {

                dis[v]=dis[u]+1;

                q.push(v);

                if(v==tp)

                    break;

            }

        }

    }

    return dis[tp]!=-1;

}

ll dfs(int u,ll cap)//寻找增广路

{

    if(u==tp||cap==0)

        return cap;

    ll res=0,f;

    for(int &i=cur[u];i!=-1;i=mp[i].next)

    {

        int v=mp[i].v;

        if(dis[v]==dis[u]+1&&(f=dfs(v,min(cap-res,mp[i].cap)))>0)

        {

            mp[i].cap-=f;

            mp[i^1].cap+=f;

            res+=f;

            if(res==cap)

                return cap;

        }

    }

    if(!res)

        dis[u]=-1;

    return res;

}

ll dinic()

{

    ll ans=0;

    while(bfs())

    {

        for(int i=0;i<=tp;i++)

            cur[i]=pre[i];

        ans+=dfs(sp,inf);

    }

    return ans;

}

int d[MAXN],a[MAXN];

int g[400][1100];

int st[400][1100];

int main() {

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        init();

        memset(d,0, sizeof(d));

        int sum=0;

        int x;

        for (int i = 1; i <=m ; ++i) {

            scanf("%d",&x);

            add(n+i,n+m+1,inf);

            d[n+i]-=x;

            d[n+m+1]+=x;

        }

        int b;

        for (int i = 1; i <=n ; ++i) {

            scanf("%d%d",&a[i],&b);

            add(0,i,b);

            int l,r;

            for (int j = 0; j <a[i] ; ++j) {

                scanf("%d%d%d",&x,&l,&r);

                add(i,n+x+1,r-l);

                d[i]-=l;

                d[n+x+1]+=l;

                st[i][j]=l;

                g[i][j]=(cnt-2);

            }

        }

        add(n+m+1,0,inf);

        for (int i = 0 ; i <=n+m+1 ; ++i) {

            if(d[i]>0)

            {

                add(n+m+2,i,d[i]);

                sum+=d[i];

            }

            if(d[i]<0)

            {

                add(i,n+m+3,-d[i]);

            }

        }

        sp=n+m+2;tp=n+m+3;

        if(sum!=dinic())

        {

            printf("-1\n");

        }

        else

        {

            sp=0,tp=n+m+1;

            printf("%lld\n",dinic());

            for (int i = 1; i <=n ; ++i) {

                for (int j = 0; j <a[i] ; ++j) {

                    printf("%lld\n",mp[g[i][j]+1].cap+(ll)st[i][j]);

                }

            }

        }

        printf("\n");

    }

    return 0;

}

//zoj3229

//loj116