題意：一個人要給女孩子們拍照，一共 n 天，m 個女孩子，每天他至多拍 d[i] 張照片，每個女孩子總共要被至少拍 g[i] 次。在第 i 天，可以拍 c[i] 個女孩子，c[i] 個女孩子中每個女孩子在當天被拍的次數是 [li,ri]，求最多可以拍多少張照片，以及每天每個可以拍的女孩子被拍了多少張照片。

析：一個很明顯網絡流，很容易建圖，建立一個源點 s 和匯點 t，然後 s 向每一天連一條邊，容量上界上 d[i]，每個女孩子向匯點連一邊，下界是 g[i]， 上界是無窮大，然後對於每一天拍的從每一天那個結點向那個女孩子連一條容量上界是 ri，下界是 li，的邊，然後就建好圖了，就是求最大流，但是這個不是普通的網絡流，我們要把它進行轉換，首先先根據無源無匯的，建立，因為是有源匯的，所以連一條邊 t -> s，INF，然後求一次 超級源點匯點（註意不是 s t） S T 最大流，如果滿流就是有可行流，然後再求 s->t 的最大流，結果就是答案。然後再輸出解，輸出解的時候就是流再加上下界。

代碼如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1500 + 50;
const int maxm = 1e6 + 10;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }

struct Edge{
  int from, to, cap, flow;
};

struct Dinic{
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  bool vis[maxn];
  int d[maxn];
  int cur[maxn];
  
  void init(int n){
    FOR(i, n, 0)  G[i].cl;
    edges.cl;
  }
  
  void addEdge(int from, int to, int cap){
    edges.pb((Edge){from, to, cap, 0});
    edges.pb((Edge){to, from, 0, 0});
    m = edges.sz;
    G[from].pb(m - 2);
    G[to].pb(m - 1); 
  }
  
  bool bfs(){
    ms(vis, 0);  d[s] = 0;  vis[s] = 1;
    queue<int> q;  q.push(s);
    
    while(!q.empty()){
      int u = q.front();  q.pop();
      for(int i = 0; i < G[u].sz; ++i){
        Edge &e = edges[G[u][i]];
        if(!vis[e.to] && e.cap > e.flow){
          vis[e.to] = 1;
          d[e.to] = d[u] + 1;
          q.push(e.to);
        }
      }
    }
    return vis[t];
  }
  
  int dfs(int u, int a){
    if(u == t || a == 0)  return a;
    int flow = 0, f;
    for(int &i = cur[u]; i < G[u].sz; ++i){
      Edge &e = edges[G[u][i]];
      if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
        e.flow += f;
        edges[G[u][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(a == 0)  break;
      }
    }
    return flow;
  }
  
  int maxFlow(int s, int t){
    this->s = s; this->t = t;
    int flow = 0;
    while(bfs()){ ms(cur, 0);  flow += dfs(s, INF); }
    return flow;
  }
};
Dinic dinic;

int down[100000], in[100000], d[maxn], g[maxn];

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    int s = 0, t = n + m + 1;
    dinic.init(t + 10);
    ms(in, 0);
    for(int i = 1; i <= m; ++i)
      scanf("%d", g + i);
    
    int cnt = 0;
    for(int i = 1; i <= n; ++i){
      int c;  scanf("%d %d", &c, d + i);
      int x, l, r;
      while(c--){
        scanf("%d %d %d", &x, &l, &r);
        down[cnt++] = l;
        dinic.addEdge(i, x + n + 1, r - l);
        in[i] -= l;
        in[x+n+1] += l;
      }
    }
    for(int i = 1; i <= n; ++i)
      dinic.addEdge(s, i, d[i]);
    for(int i = 1; i <= m; ++i){
      dinic.addEdge(i + n, t, INF - g[i]);
      in[i+n] -= g[i];
      in[t] += g[i];
    }
    int S = t + 1, T = t + 2;
    int ans = 0;
    for(int i = 0; i <= t; ++i)
      if(in[i] > 0)  dinic.addEdge(S, i, in[i]), ans += in[i];
      else if(in[i] < 0)  dinic.addEdge(i, T, -in[i]);
    dinic.addEdge(t, s, INF);
    if(ans != dinic.maxFlow(S, T)){ printf("-1\n\n");  continue; }
    printf("%d\n", dinic.maxFlow(s, t));
    for(int i = 0; i < cnt; ++i)  printf("%d\n", dinic.edges[i<<1].flow + down[i]);
    printf("\n");
  }
  return 0;
}

ZOJ 3229 Shoot the Bullet (有源有匯有上下界最大流)