UVALive - 2957 Bring Them There(最大流 图论建模)

题目大意：有n个星球，你的任务是用最短的时间把k个超级计算机从S星球运送到T星球。每个超级计算机需要一艘飞船来运输，行星之间有m条双向隧道，每条隧道需要一天时间来通过，且不能有两艘飞船同时使用同一条隧道，隧道不会连接两个相同的行星，且每一对行星之间最多只有一条隧道

解题思路：按照大白书上的思路是拆点
比如运送的时间为T，那么就把每个点u拆成T + 1个，分别为u0, u1 … uT，分别对应的是第i天的u星球
对于所给的每条边(u,v)，假设是第i天，就连边ui –>vi+1和vi–>ui+1，两条边的容量都为1
还得添加一个点ui—>ui+1，容量为INF，表示飞船该天没有前行
计算过程的话，就逐步扩大图，一天一天的枚举，一天一天的扩大，直到流量为k为止
这里的话得注意一点，在某时刻a–>bi+1和bi–>ai+1同时有流量是不允许的，这里的话，如果出现了两个方向都有流量了，那我们就相当于把这两个飞船分别停留在a星球和b星球上各一天，这样就不会矛盾了

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;#define M 1000010#define N 10010 #define INF 0x3f3f3f3fstruct Edge{    int u, v, cap, flow, next;}E[M];struct Dinic{    int head[N], d[N];    int tot, sink, source;    void init() {        memset(head, -1, sizeof(head));        tot = 0;    }    inline void AddEdge(int u, int v, int cap) {        E[tot].u = u; E[tot].v = v; E[tot].cap = cap; E[tot].flow = 0; E[tot].next = head[u]; head[u] = tot++;        u = u ^ v; v = u ^ v; u = u ^ v;        E[tot].u = u; E[tot].v = v; E[tot].cap = 0; E[tot].flow = 0; E[tot].next = head[u]; head[u] = tot++;    }    inline bool bfs(int s) {        int u, v;        memset(d, 0, sizeof(d));        queue<int> Q;        Q.push(s);        d[s] = 1;        while (!Q.empty()) {            u = Q.front(); Q.pop();            if (u == sink) return true;            for (int i = head[u]; ~i; i = E[i].next) {                v = E[i].v;                if (!d[v] && E[i].cap - E[i].flow > 0) {                    d[v] = d[u] + 1;                    Q.push(v);                }            }        }        return false;    }    int dfs(int x, int a) {        if (x == sink || a == 0)            return a;        int f, flow = 0;        for (int i = head[x]; ~i; i = E[i].next) {            int v = E[i].v;            if (d[v] == d[x] + 1 && E[i].cap - E[i].flow > 0) {                f = dfs(v, min(a, E[i].cap - E[i].flow));                E[i].flow += f;                E[i^1].flow -= f;                flow += f;                a -= f;                if (!a) break;            }        }        if (flow == 0) d[x] = 0;        return flow;    }    int Maxflow(int source, int sink, int need) {        int flow = 0;        this->sink = sink;        while (bfs(source)) {            flow += dfs(source, need - flow);            if (flow >= need) return flow;        }        return flow;    }};Dinic dinic;struct Node {    int x, y;}node[N];int n, m, k, s, t;int pos[N];bool vis[N];void print(int cnt) {    for (int i = 1; i <= k; i++) pos[i] = s;    printf("%d\n", cnt);    for (int i = 1; i <= cnt; i++) {        vector<int> u, v;        memset(vis, 0, sizeof(vis));        for (int j = 0; j < 4 * m; j += 4) {            int p = j + (i - 1) * 4 * m + n * i * 2;            //正向有流量，反向没流量，也就是u-->v            if (E[p].flow && !E[p+2].flow) {                u.push_back(E[p ^ 1].v - (i - 1) * n);                v.push_back(E[p].v - i * n);            }//反向有流量，正向没流量，也就是v-->u,这样的话就排除了两边都有流量的情况，相当于两点交换了，也就是两艘飞船待在原地了            else if (!E[p].flow && E[p + 2].flow) {                u.push_back(E[(p + 2) ^ 1].v - (i - 1) * n);                v.push_back(E[p + 2].v - i * n);            }        }        printf("%d", u.size());        for (int p = 0; p < u.size(); p++)            for (int j = 1; j <= k; j++) {                if (!vis[j] && pos[j] == u[p]) {                    vis[j] = 1;                    printf(" %d %d", j, v[p]);                    pos[j] = v[p];                    break;                }            }        printf("\n");    }}void solve() {    for (int i = 0; i < m; i++) scanf("%d%d", &node[i].x, &node[i].y);    int cnt = 0, Maxflow = 0, sink = t;    dinic.init();    while (Maxflow < k) {        ++cnt;        //待在原地        for (int i = 1; i <= n; i++) dinic.AddEdge(i + (cnt - 1) * n,  i + cnt * n, INF);        //边        for (int i = 0; i < m; i++) {            dinic.AddEdge(node[i].x + (cnt - 1) * n, node[i].y + cnt * n, 1);            dinic.AddEdge(node[i].y + (cnt - 1) * n, node[i].x + cnt * n, 1);        }        sink += n;        Maxflow += dinic.Maxflow(s, sink, k - Maxflow);    }    print(cnt);}int main() {    while (scanf("%d%d%d%d%d", &n, &m, &k, &s, &t) != EOF) solve();    return 0;}