POJ 3204 Ikki's Story I - Road Reconstruction
題目連結
題意:給定一個有向圖,求出最大流後,問哪些邊增加容量後,可以使最大流增加
思路:對于一個可以增加的,必然原來就是滿流,并且從源點到彙點,的一條路徑上,都是還有殘留容量的,這樣隻要從源點和彙點分别出發dfs一遍,标記掉經過點,然後枚舉滿流邊,如果兩端都是标記過的點,這個邊就是可以增加的
代碼:
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXNODE = 505;
const int MAXEDGE = 100005;
typedef int Type;
const Type INF = 0x3f3f3f3f;
struct Edge {
int u, v;
Type cap, flow;
int cnt;
Edge() {}
Edge(int u, int v, Type cap, Type flow) {
this->u = u;
this->v = v;
this->cap = cap;
this->flow = flow;
this->cnt = 0;
}
};
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) {
edges[m] = Edge(u, v, cap, 0);
next[m] = first[u];
first[u] = m++;
edges[m] = Edge(v, u, 0, 0);
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])
cut.push_back(i);
}
}
bool mark[MAXNODE][2];
void find(int u, int tp) {
mark[u][tp] = true;
for (int i = first[u]; i + 1; i = next[i]) {
int v = edges[i].v;
if (mark[v][tp]) continue;
if (tp == 0 && i % 2) continue;
if (tp && i % 2 == 0) continue;
if (edges[i^tp].cap == edges[i^tp].flow) continue;
find(v, tp);
}
}
int solve() {
Maxflow(0, n - 1);
memset(mark, false, sizeof(mark));
find(0, 0);
find(n - 1, 1);
int ans = 0;
for (int i = 0; i < m; i += 2)
if (edges[i].cap == edges[i].flow && mark[edges[i].u][0] && mark[edges[i].v][1]) ans++;
return ans;
}
} gao;
int n, m;
int main() {
while (~scanf("%d%d", &n, &m)) {
gao.init(n);
int u, v, w;
while (m--) {
scanf("%d%d%d", &u, &v, &w);
gao.add_Edge(u, v, w);
}
printf("%d\n", gao.solve());
}
return 0;
}