POJ 3204 Ikki's Story I - Road Reconstruction（最大流）

POJ 3204 Ikki's Story I - Road Reconstruction

题目链接

题意：给定一个有向图，求出最大流后，问哪些边增加容量后，可以使最大流增加

思路：对于一个可以增加的，必然原来就是满流，并且从源点到汇点，的一条路径上，都是还有残留容量的，这样只要从源点和汇点分别出发dfs一遍，标记掉经过点，然后枚举满流边，如果两端都是标记过的点，这个边就是可以增加的

代码：

#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;
}