### 我出过的东方系列题目 » ZOJ3229

ZOJ3229.cpp

```#include <queue>
#include <cstdio>
#include <vector>
#include <cassert>
#include <algorithm>

using namespace std;

const int MAXN = 365 + 1000 + 10;
const int MAXM = 100000;
inline int RE(int i) { return i ^ 1; }

struct Edge {
int v;
int c;
};

struct NoSuchFlowException {
};

struct ExtFlowNetwork {
int n, m, source, sink;
vector<int> e[MAXN];
Edge edge[MAXM * 2];

int diff[MAXN];		//*

void init(int n, int source, int sink) {	// if no source & sink, then source = sink = -1
this->n = n;
this->m = 0;
this->source = source;
this->sink = sink;
for (int i = 0; i < n; ++i) {
e[i].clear();
}
fill(diff, diff + n, 0);
}

void _addEdge(int a, int b, int c) {
edge[m].v = b;
edge[m].c = c;
e[a].push_back(m);
edge[m + 1].v = a;
edge[m + 1].c = 0;
e[b].push_back(m + 1);
m += 2;
}

void addEdge(int a, int b, int l, int u) {
diff[a] += l;
diff[b] -= l;
}

// shortest augmenting path
int c[MAXN];		//*
int d[MAXN];		//*
int done[MAXN];		//*
int path[MAXN];		//*
int len;			//*

void _bfs() {
queue<int> q;
fill(c, c + n, 0);
fill(d, d + n, n);	// n->inf
d[sink] = 0;
q.push(sink);
while (!q.empty()) {
int cur = q.front();
++c[d[cur]];
for (size_t i = 0; i < e[cur].size(); ++i) {
int id = e[cur][i];
if (d[edge[id].v] == n && edge[RE(id)].c > 0) {
d[edge[id].v] = d[cur] + 1;
q.push(edge[id].v);
}
}
q.pop();
}
}

void _retreat(int v) {
--c[d[v]];
d[v] = n;	// n->inf
for (size_t i = 0; i < e[v].size(); ++i) {
Edge &arc = edge[e[v][i]];
if (d[v] > d[arc.v] + 1 && arc.c > 0) {
d[v] = d[arc.v] + 1;
done[v] = i;	// ~
}
}
++c[d[v]];
}

int _augment() {
int todo = -1;
int flow = 0x7fffff;
for (int i = 0; i < len; ++i) {
Edge &arc = edge[e[path[i]][done[path[i]]]];
if (arc.c < flow) {
flow = arc.c;
todo = i;
}
}
for (int i = 0; i < len; ++i) {
int id = e[path[i]][done[path[i]]];
edge[id].c -= flow;
edge[RE(id)].c += flow;
}
len = todo;
return flow;
}

int sap() {
int flow = 0;
_bfs();
fill(done, done + n, 0);
len = 0;
path[0] = source;
while (d1 != n/* && len >= 0*/) {	// n->inf
int back = path[len];
if (back == sink) {
flow += _augment();
} else {
while (done[back] < (int)e[back].size()) {
Edge &arc = edge[e[back][done[back]]];
if (d[arc.v] == d[back] - 1 && arc.c > 0) {
break;
} else {
++done[back];
}
}
if (done[back] == (int)e[back].size()) {
if (c[d[back]] == 1) {
break;	// return;
} else {
_retreat(back);
if (back != source) {	// !!
--len;	// pop
}
}
} else {
path[++len] = edge[e[back][done[back]]].v;	// push
}
}
}
return flow;
}
// end of sap

void feasibleFlow() {	// throws NoSuchFlowException
int N = n + 2;
int M = m;
int SOURCE = n;
int SINK = n + 1;
int flow = 0;
e[SOURCE].clear();
e[SINK].clear();
if (source != sink/* != -1*/) {
}
for (int i = 0; i < n; ++i) {
if (diff[i] > 0) {
flow += diff[i];
} else/* if (diff[i] <= 0)*/ {	// an edge is added even if diff[i] == 0 for [@1]
}
}
swap(n, N);
swap(source, SOURCE);
swap(sink, SINK);
int aflow = sap();
swap(n, N);
swap(source, SOURCE);
swap(sink, SINK);
for (int i = 0; i < n; ++i) {
e[i].pop_back();	// assert(...) :[@1]
}
if (source != sink/* != -1*/) {
e1.pop_back();
e[sink].pop_back();
}
// e[SOURCE].clear();
// e[SINK].clear();
m = M;
if (aflow != flow) {
throw NoSuchFlowException();
}
}

void minFlow() {
feasibleFlow();
swap(source, sink);
sap();
swap(source, sink);
}

void maxFlow() {
feasibleFlow();
sap();
}
};

#define N() (n + m + 2)	// number
#define S() (n + m)		// source
#define T() (n + m + 1)	// sink
#define D(i) (i)		// input layer: day i
#define G(j) (n + j)	// output layer: girl j

const int INF = 1000000000;

int main() {
int n, m, p, c, t, l;
int d[365], g[1000], r[365 * 100];
ExtFlowNetwork efn;

while (scanf("%d%d", &n, &m) != EOF) {
assert(1 <= n && n <= 365);				/// @param	n	1 <= n <= 365
assert(1 <= m && m <= 1000);			/// @param	m1 <= m <= 1000
efn.init(N(), S(), T());
for (int i = 0; i < m; ++i) {
assert(scanf("%d", &g[i]) == 1);
assert(0 <= g[i] && g[i] <= 10000);	/// @param	gG in range [0, 10000]
}
p = 0;
for (int i = 0; i < n; ++i) {
assert(scanf("%d%d", &c, &d[i]) == 2);
assert(1 <= c && c <= 100);			/// @param	c	1 <= C <= 100
assert(0 <= d[i] && d[i] <= 30000);	/// @param	d0 <= D <= 30000
for (int j = 0; j < c; ++j) {
assert(scanf("%d%d%d", &t, &l, &r[p]) == 3);
assert(0 <= t && t < m);		/// @param	0 <= T < m
assert(0 <= l && l <= r[p] && r[p] <= 100);	/// 0 <= T < m, 0 <= L <= R <= 100.
++p;
}
}
for (int i = 0; i < n; ++i) {
}
for (int i = 0; i < m; ++i) {
}
try {
efn.maxFlow();
int ans = 0;
for (int i = 0; i < p; ++i) {
ans += r[i] - efn.edge[i << 1].c;
}
printf("%d\n", ans);
//
for (vector<int>::const_iterator it = efn.e[S()].begin(); it != efn.e[S()].end(); ++it) {
ans -= efn.edge[*it ^ 1].c;
}
assert(ans == 0);
//
for (int i = 0; i < p; ++i) {
printf("%d\n", r[i] - efn.edge[i << 1].c);
}
} catch (NoSuchFlowException e) {
puts("-1");
}
puts("");	/// Output a blank line after each case.
}

return 0;
}

// vim: ft=cpp.doxygen sw=4 ts=4
```
9 Responses to “ZOJ3229”
1. niao says:

请问下您，为什么我这道题TLE了。。我的SAP模版经过细心优化了，以前的网络流跑在几乎前面

2. kkkk says:

oh, sorry,submit the same context many times……..

• xiaodao says:

… … Maybe I could explain this point to you …

This is a “模板” of network_flow with “上下界” ..
It could solve both 最小流 && 最大流 …

The case that ans < 0 .. could happend in the 最小流 .. which is not truely happend in this problem …

http://acm.sgu.ru/

3. kkkk says:

why i use your extFlowNetwork to slove sgu 176 but always Wrong Answer on test 12

• watashi says:

I got this problem accepted. Remember that “with the non-negative speed”. If you simply use efn.minFlow(), you may get a negative answer.

• kkkk says:

thanks,
when the ans < 0, add a new source,add an edge between new source and the old source whose capacity is -ans, is method true?
Could you tell me your solution when ans < 0.
Thanks

• watashi says:

I add a new source 0, and then add an edge from 0 to 1 whose lower bound is 0 and upper bound is infinity so that the efn.minFlow() will alway be between 0 and infinity, namely non-negative.

```	efn.init(n + 1, 0, n);
for (int i = 0; i < m; ++i) {
scanf("%d%d%d%d", &u, &v, &z[i], &c);
efn.addEdge(u, v, c * z[i], z[i]);
}
```
• kkkk says:

thanks,
when the ans < 0, add a new source,add an edge between new source and the old source whose capacity is -ans, the sap() again, is this method true?
Could you tell me your solution when ans < 0.
Thanks

• kkkk says:

thanks,
when the ans < 0, add a new source,add an edge between new source and the old source whose capacity is -ans, then sap() again, is this method true?
Could you tell me your solution when ans < 0.
Thanks

4.