HDU - 3251 Being a Hero(最小割)

题目大意：有一个国王要奖励英雄一些城镇，英雄可以从国王奖励的城镇中挑选k个，每个城镇都有相应的value
但国王又不想走到这个英雄选择的城镇，所以英雄选择了城镇的时候就要考虑要炸掉哪些路，使得国王走不到他的城镇，每条道路都有相应的value
问英雄的最大获利，和需要炸掉的路是哪些

解题思路：假设国王所在的地点是u，英雄所选择的城镇是v，现在的任务就是破坏掉s–>u…->v->t(s是源点，t是汇点)且花费的价值最小，所以就是要求最小割了
剩下的任务就是建图了,源点和国王所在地连接，容量为INF
英雄可以选择的城镇跟汇点连接，容量为相应的value
剩下的就是有向道路的连接了
最大的价值就是可选的城镇的总value - 最小割(最大流)

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;const int MAXNODE = 1010;const int MAXEDGE = 300010;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge{    int u, v, next, id;    Type cap, flow;    Edge() {}    Edge(int u, int v, Type cap, Type flow, int next, int id) : u(u), v(v), cap(cap), flow(flow), next(next), id(id){}};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, int id) {        edges[m] = Edge(u, v, cap, 0, head[u], id);        head[u] = m++;        edges[m] = Edge(v, u, 0, 0, head[v], id);        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] && edges[i].id != 0)                 cut.push_back(i);        }    }}dinic;int n, m, f, cas = 1;void solve() {    scanf("%d%d%d", &n, &m, &f);    int source = 0, sink = n + 1;    dinic.init(sink + 1);    dinic.AddEdge(source, 1, INF, 0);    int u, v, c;    int Sum = 0;    for (int i = 1; i <= m; i++) {        scanf("%d%d%d", &u, &v, &c);        dinic.AddEdge(u, v, c, i);    }    for (int i = 1; i <= f; i++) {        scanf("%d%d", &u, &c);        Sum += c;        dinic.AddEdge(u, sink, c, 0);    }    printf("Case %d: %d\n", cas++, Sum - dinic.Maxflow(source, sink));    dinic.Mincut();    printf("%d", dinic.cut.size());    for (int i = 0; i < dinic.cut.size(); i++)         printf(" %d", dinic.edges[dinic.cut[i]].id);    printf("\n");}int main() {    int test;    scanf("%d", &test);    while (test--) solve();    return 0;}