传送
题面:给定一个有向图,每条边均有一个容量。问是否存在一个从点\(1\)到点\(n\),流量为\(c\)的流。如果不存在,是否可以恰好修改一条边的容量,使得存在这样的流?
首先如果最大流大于等于\(C\),那直接输出possible;
否则要修改的一条边一定在最小割上,所以我们只要依次将每条最小割边的容量改成\(C\),再跑最大流检查即可。
但这样最多要跑\(m\)次DInic,会TLE,所以这题的重点在于以下两点优化:
- 第一次求完最大流后把流量留着,以后都在这个基础上增广。
- 每次不用求最大流,大于等于\(C\)就可以返回。
代码细节,还是有必要说一下的:
- 判断是否是最小割边:
for(int i = 1; i <= n; ++i) forE(j, i, v) //访问每条边
if(dis[i] && !dis[v] && e[j].cap > 0) //刚好在残量网络的断边上
ret.emplace_back(i, j);
因为Dinic在bfs的时候只走流量未满的边,所以如果有的边满流量,且刚好位于残量网络能延伸的尽头,那么这些边就构成了最小割。
2. 在第一次求完最大流的基础上增广:
for(int i = 0; i <= ecnt; ++i) e[i].cap -= e[i].flow;
即改变每条边的流量,这样就能记住当前的残量网络了。
3. 改每一条割边后不求最大流,大于等于\(C\)就返回:
for(int i = 0; i < (int)cuts.size(); ++i)
{
int x = cuts[i].S;
e[x].cap = C; //将流量修改为C
clearFlow();
if(flow + maxFlow(C - flow) >= C) ans.emplace_back(cuts[i].F, e[x].to);
e[x].cap = 0; //改回来,因为是割边,所以必满流,新的容量就是0
}
以及求最大流的部分:
In int maxFlow(int lim)
{
int flow = 0;
while(bfs())
{
memcpy(cur, head, sizeof(head));
flow += dfs(s, lim - flow);
//这我没搞懂,对于这道题改成dfs(s, INF)也能AC,但对于POJ1895这道题就不行
if(flow >= lim) return flow;
}
return flow;
}
以下是完整代码:
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 105;
const int maxe = 2e4 + 5;
In ll read()
{
ll ans = 0;
char ch = getchar(), las = ' ';
while(!isdigit(ch)) las = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(las == '-') ans = -ans;
return ans;
}
In void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int n, m, C, s, t;
struct Edge
{
int nxt, to, cap, flow;
}e[maxe];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y, int w)
{
e[++ecnt] = (Edge){head[x], y, w, 0};
head[x] = ecnt;
e[++ecnt] = (Edge){head[y], x, 0, 0};
head[y] = ecnt;
}
int dis[maxn];
In bool bfs()
{
Mem(dis, 0), dis[s] = 1;
queue<int> q; q.push(s);
while(!q.empty())
{
int now = q.front(); q.pop();
for(int i = head[now], v; ~i; i = e[i].nxt)
if(e[i].cap > e[i].flow && !dis[v = e[i].to])
dis[v] = dis[now] + 1, q.push(v);
}
return dis[t];
}
int cur[maxn];
In int dfs(int now, int res)
{
if(now == t || res == 0) return res;
int flow = 0, f;
for(int& i = cur[now], v; ~i; i = e[i].nxt)
{
if(dis[v = e[i].to] == dis[now] + 1 && (f = dfs(v, min(res, e[i].cap - e[i].flow))) > 0)
{
e[i].flow += f, e[i ^ 1].flow -= f;
flow += f, res -= f;
if(res == 0) break;
}
}
return flow;
}
In int maxFlow(int lim)
{
int flow = 0;
while(bfs())
{
memcpy(cur, head, sizeof(head));
flow += dfs(s, lim - flow);
if(flow >= lim) return flow;
}
return flow;
}
#define pr pair<int, int>
#define mp make_pair
#define F first //起点
#define S second //边所在的编号
In vector<pr> minCut()
{
bfs();
vector<pr> ret;
for(int i = 1; i <= n; ++i) forE(j, i, v)
if(dis[i] && !dis[v] && e[j].cap > 0)
ret.emplace_back(i, j);
return ret;
}
In void clearFlow()
{
for(int i = 0; i <= ecnt; ++i) e[i].flow = 0;
}
vector<pr> ans;
In void solve(int flow)
{
ans.clear();
vector<pr> cuts = minCut();
for(int i = 0; i <= ecnt; ++i) e[i].cap -= e[i].flow;
for(int i = 0; i < (int)cuts.size(); ++i)
{
int x = cuts[i].S; //割边必满流
e[x].cap = C;
clearFlow();
if(flow + maxFlow(C - flow) >= C) ans.emplace_back(cuts[i].F, e[x].to);
e[x].cap = 0;
}
}
int main()
{
int T = 0;
while(scanf("%d%d%d", &n, &m, &C) && (n | m | C))
{
Mem(head, -1), ecnt = -1;
s = 1, t = n;
for(int i = 1; i <= m; ++i)
{
int x = read(), y = read(), w = read();
addEdge(x, y, w);
}
printf("Case %d: ", ++T);
int flow = maxFlow(C);
if(flow >= C) {puts("possible"); continue;}
else solve(flow);
if(ans.empty()) puts("not possible");
else
{
sort(ans.begin(), ans.end());
printf("possible option:");
for(int i = 0; i < (int)ans.size(); ++i)
printf("(%d,%d)%c", ans[i].F, ans[i].S, i == (int)ans.size() - 1 ? '\n' : ',');
}
}
return 0;
}