UVa11248 Frequency Hopping 網路擴容
阿新 • • 發佈:2021-06-18
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題面:給定一個有向圖,每條邊均有一個容量。問是否存在一個從點\(1\)到點\(n\),流量為\(c\)的流。如果不存在,是否可以恰好修改一條邊的容量,使得存在這樣的流?
首先如果最大流大於等於\(C\),那直接輸出possible;
否則要修改的一條邊一定在最小割上,所以我們只要依次將每條最小割邊的容量改成\(C\),再跑最大流檢查即可。
但這樣最多要跑\(m\)次DInic,會TLE,所以這題的重點在於以下兩點優化:
- 第一次求完最大流後把流量留著,以後都在這個基礎上增廣。
- 每次不用求最大流,大於等於\(C\)就可以返回。
程式碼細節,還是有必要說一下的:
- 判斷是否是最小割邊:
for(int i = 1; i <= n; ++i) forE(j, i, v) //訪問每條邊
if(dis[i] && !dis[v] && e[j].cap > 0) //剛好在殘量網路的斷邊上
ret.emplace_back(i, j);
因為Dinic在bfs的時候只走流量未滿的邊,所以如果有的邊滿流量,且剛好位於殘量網路能延伸的盡頭,那麼這些邊就構成了最小割。
2. 在第一次求完最大流的基礎上增廣:
for(int i = 0; i <= ecnt; ++i) e[i].cap -= e[i].flow;
即改變每條邊的流量,這樣就能記住當前的殘量網路了。
3. 改每一條割邊後不求最大流,大於等於\(C\)
for(int i = 0; i < (int)cuts.size(); ++i)
{
int x = cuts[i].S;
e[x].cap = C; //將流量修改為C
clearFlow();
if(flow + maxFlow(C - flow) >= C) ans.emplace_back(cuts[i].F, e[x].to);
e[x].cap = 0; //改回來,因為是割邊,所以必滿流,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); //這我沒搞懂,對於這道題改成dfs(s, INF)也能AC,但對於POJ1895這道題就不行 if(flow >= lim) return flow; } return flow; }
以下是完整程式碼:
#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 //起點
#define S second //邊所在的編號
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; //割邊必滿流
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;
}