ZOJ - 3229 Shoot the Bullet(有源有汇上下界最大流)

题目大意：有一个屌丝要给M个女神拍照，每个女神至少要拍G张照片
给出屌丝每天的拍照任务，每天要给C个女神拍照，因为内存有限，所以每天至多拍D张照片
女神i要求该天至少要拍L张，至多拍R张
问屌丝能否满足所有的女神的要求，因为是屌丝，当然希望拍的照片越多越好了，所以，要求求出这个屌丝能拍的最多张照片是多少张

解题思路：有源有汇上下界最大流，将每个女神抽象成一个点，每天抽象成一个点，每个任务抽象成一条边，建图
源点连接每一天，下界为0，上界为D
汇点连接每个女神，下界为G，上界为INF
每一天和该天任务的女神连线，下界为L，上界为R

附上一个有讲解上下界最大流的传送门

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXNODE = 1400;
const int MAXEDGE = 1000010;
typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge{
    int u, v, next;
    Type cap, flow;
    Edge() {}
    Edge(int u, int v, Type cap, Type flow, int next) : u(u), v(v), cap(cap), flow(flow), next(next){}
};

struct Dinic{
    int n, m, s, t;
    Edge edges[MAXEDGE];
    int head[MAXNODE];
    int cur[MAXNODE];
    bool vis[MAXNODE];
    Type d[MAXNODE];
    vector<int> cut;

    void init(int n) {
        this->n = n;
        memset(head, -1, sizeof(head));
        m = 0;
    }

    void AddEdge(int u, int v, Type cap) {
        edges[m] = Edge(u, v, cap, 0, head[u]);
        head[u] = m++;
        edges[m] = Edge(v, u, 0, 0, head[v]);
        head[v] = m++;
    } 

    bool BFS() {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        Q.push(s);
        d[s] = 0;
        vis[s] = 1;

        while (!Q.empty()) {
            int u = Q.front(); Q.pop();
            for (int i = head[u]; ~i; i = edges[i].next) {
                Edge &e = edges[i];
                if (!vis[e.v] && e.cap > e.flow) {
                    vis[e.v] = true;
                    d[e.v] = d[u] + 1;
                    Q.push(e.v);
                }
            }
        }
        return vis[t];
    }

    Type DFS(int u, Type a) {
        if (u == t || a == 0) return a;

        Type flow = 0, f;
        for (int &i = cur[u]; i != -1; i = edges[i].next) {
            Edge &e = edges[i];
            if (d[u] + 1 == d[e.v] && (f = DFS(e.v, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[i ^ 1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }

    Type Maxflow(int s, int t) {
        this->s = s; this->t = t;
        Type flow = 0;
        while (BFS()) {
            for (int i = 0; i < n; i++)
                cur[i] = head[i];
            flow += DFS(s, INF);
        }
        return flow;
    }

    void Mincut() {
        cut.clear();
        for (int i = 0; i < m; i += 2) {
            if (vis[edges[i].u] && !vis[edges[i].v]) 
                cut.push_back(i);
        }
    }

}dinic;

#define maxn 2010
int source, sink, SuperSource, SuperSink, n, m;
int d[maxn], ans[MAXEDGE], MaxDay[maxn];

void solve() {
    source = n + m, sink = source + 1;
    SuperSource = sink + 1, SuperSink = SuperSource + 1;
    dinic.init(SuperSink + 1);
    memset(d, 0, sizeof(d));

    int G;
    for (int i = 0; i < m; i++) {
        scanf("%d", &G);
        d[sink] += G;
        d[i + n] -= G;
        dinic.AddEdge(i + n, sink, INF);
    }

    int start = dinic.m;
    int cnt = 0, T, L, R, C;
    for (int i = 0; i < n; i++) {
        scanf("%d%d", &C, &MaxDay[i]);
        for (int j = 0; j < C; j++) {
            scanf("%d%d%d", &T, &L, &R);
            d[i] -= L;
            d[T + n] += L;
            ans[cnt] = L;
            dinic.AddEdge(i, T + n, R - L);
            cnt++;
        }
    }
    int end = dinic.m;

    for (int i = 0; i < n; i++) 
        dinic.AddEdge(source, i, MaxDay[i]);
    int Sum = 0;
    for (int i = 0; i <= sink; i++) {
        if (d[i] > 0) {
            Sum += d[i];
            dinic.AddEdge(SuperSource, i, d[i]);
        }
        if (d[i] < 0) dinic.AddEdge(i, SuperSink, -d[i]);
    }

    dinic.AddEdge(sink, source, INF);
    if (dinic.Maxflow(SuperSource, SuperSink) != Sum) printf("-1\n");
    else {
        dinic.head[SuperSource] = -1;
        dinic.head[SuperSink] = -1;
        int flow = dinic.Maxflow(source, sink);
        printf("%d\n", dinic.edges[dinic.m - 2].flow + flow);
        for (int i = start, j = 0; i < end; i += 2, j++)
            printf("%d\n", ans[j] + dinic.edges[i].flow);
    }
    printf("\n");
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF) solve();
    return 0;
}