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HDU 3215 Being a Hero(最小割)
题意:一个英雄,分到几个城市,每个城市有一个价值,但是要求分到城市后,必须破坏掉道路使得首都1都不能到达,破坏道路有开销,问最大能获得的收益和需要破坏的道路ID
思路:最小割,城市1做源点,有向边建图,容量为代价,然后每个可以分的城市连到汇点,容量为价值,跑一下最小割即可
代码:
[cpp] view plaincopy
- #include <cstdio>
- #include <cstring>
- #include <queue>
- #include <algorithm>
- using namespace std;
- const int MAXNODE = 1005;
- const int MAXEDGE = 300005;
- typedef int Type;
- const Type INF = 0x3f3f3f3f;
- struct Edge {
- int u, v, id;
- Type cap, flow;
- Edge() {}
- Edge(int u, int v, Type cap, Type flow, int id) {
- this->u = u;
- this->v = v;
- this->cap = cap;
- this->flow = flow;
- this->id = id;
- }
- };
- struct Dinic {
- int n, m, s, t;
- Edge edges[MAXEDGE];
- int first[MAXNODE];
- int next[MAXEDGE];
- bool vis[MAXNODE];
- Type d[MAXNODE];
- int cur[MAXNODE];
- vector<int> cut;
- void init(int n) {
- this->n = n;
- memset(first, -1, sizeof(first));
- m = 0;
- }
- void add_Edge(int u, int v, Type cap, int id) {
- edges[m] = Edge(u, v, cap, 0, id);
- next[m] = first[u];
- first[u] = m++;
- edges[m] = Edge(v, u, 0, 0, id);
- next[m] = first[v];
- first[v] = m++;
- }
- bool bfs() {
- memset(vis, false, sizeof(vis));
- queue<int> Q;
- Q.push(s);
- d[s] = 0;
- vis[s] = true;
- while (!Q.empty()) {
- int u = Q.front(); Q.pop();
- for (int i = first[u]; i != -1; i = next[i]) {
- 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 = next[i]) {
- 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] = first[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)
- cut.push_back(i);
- }
- int sz = cut.size();
- printf("%d", sz);
- for (int i = 0; i < sz; i++)
- printf(" %d", edges[cut[i]].id);
- printf("\n");
- }
- } gao;
- int t, n, m, f;
- int main() {
- int cas = 0;
- scanf("%d", &t);
- while (t--) {
- scanf("%d%d%d", &n, &m, &f);
- gao.init(n + 1);
- int u, v, w, tot = 0;
- for (int i = 1; i <= m; i++) {
- scanf("%d%d%d", &u, &v, &w);
- gao.add_Edge(u, v, w, i);
- }
- while (f--) {
- scanf("%d%d", &u, &w);
- tot += w;
- gao.add_Edge(u, 0, w, 0);
- }
- printf("Case %d: %d\n", ++cas, tot - gao.Maxflow(1, 0));
- gao.MinCut();
- }
- return 0;
- }
HDU 3215 Being a Hero(最小割)
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