題目鏈接:HDU 1532 Drainage Ditches 最大排水量
Drainage Ditches
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9641 Accepted Submission(s): 4577
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
分析:最大流,EK算法。
代碼:
EK:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define maxn 220
#define INF 0x3f3f3f3f
int ans, s, t, n;
int a[maxn], pre[maxn];
int flow[maxn][maxn];
int cap[maxn][maxn];
void Edmonds_Karp()
{
queue<int> q;
memset(flow, 0, sizeof(flow));
ans = 0;
while(1)
{
memset(a, 0, sizeof(a));
a[s] = INF;
q.push(s);
while(!q.empty()) //bfs找增廣路徑
{
int u = q.front();
q.pop();
for(int v = 1; v <= n; v++)
if(!a[v] && cap[u][v] > flow[u][v])
{
pre[v] = u;
q.push(v);
a[v] = min(a[u], cap[u][v]-flow[u][v]);
}
}
if(a[t] == 0) break;
for(int u = t; u != s; u = pre[u]) //改進網絡流
{
flow[pre[u]][u] += a[t];
flow[u][pre[u]] -= a[t];
}
ans += a[t];
}
}
int main()
{
//freopen("hdu_1532.txt", "r", stdin);
int m, u, v, c;
while(~scanf("%d%d", &m, &n))
{
memset(cap, 0, sizeof(cap));
while(m--)
{
scanf("%d%d%d", &u, &v, &c);
cap[u][v] += c;
}
s = 1, t = n;
Edmonds_Karp();
printf("%d\n", ans);
}
return 0;
}
Dinic:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define maxn 220
#define INF 0x3f3f3f3f
struct Edge
{
int from, to, cap;
};
vector<Edge> EG;
vector<int> G[maxn];
int n, s, t, ans, d[maxn], cur[maxn];
void addEdge(int from, int to, int cap)
{
EG.push_back((Edge){from, to, cap});
EG.push_back((Edge){to, from, 0});
int x = EG.size();
G[from].push_back(x-2);
G[to].push_back(x-1);
}
bool bfs()
{
memset(d, -1, sizeof(d));
queue<int> q;
q.push(s);
d[s] = 0;
while(!q.empty())
{
int x = q.front();
q.pop();
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(d[e.to] == -1 && e.cap > 0)
{
d[e.to] = d[x]+1;
q.push(e.to);
}
}
}
return (d[t]!=-1);
}
int dfs(int x, int a)
{
if(x == t || a == 0) return a;
int flow = 0, f;
for(int& i = cur[x]; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(d[x]+1 == d[e.to] && (f = dfs(e.to, min(a, e.cap))) > 0)
{
e.cap -= f;
EG[G[x][i]^1].cap += f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
void Dinic()
{
ans = 0;
while(bfs())
{
memset(cur, 0, sizeof(cur));
ans += dfs(s, INF);
}
}
int main()
{
//freopen("hdu_1532.txt", "r", stdin);
int m, u, v, c;
while(~scanf("%d%d", &m, &n))
{
while(m--)
{
scanf("%d%d%d", &u, &v, &c);
addEdge(u, v, c);
}
s = 1, t = n;
Dinic();
printf("%d\n", ans);
EG.clear();
for(int i = 0; i <= n; ++i)
G[i].clear();
}
return 0;
}
(3)ISAP:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
#define maxn 203
#define INF 0x3f3f3f3f
struct Edge
{
int from, to, cap, flow;
};
vector<Edge> EG;
vector<int> G[maxn];
int n, s, t, ans, d[maxn], cur[maxn], p[maxn], num[maxn];
bool vis[maxn];
void addEdge(int from, int to, int cap)
{
EG.push_back((Edge){from, to, cap, 0});
EG.push_back((Edge){to, from, 0, 0});
int x = EG.size();
G[from].push_back(x-2);
G[to].push_back(x-1);
}
void bfs()
{
memset(vis, false, sizeof(vis));
queue<int> q;
vis[t] = true;
d[t] = 0;
q.push(t);
while(!q.empty())
{
int x = q.front();
q.pop();
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]^1];
if(!vis[e.from] && e.cap > e.flow)
{
vis[e.from] = true;
d[e.from] = d[x]+1;
q.push(e.from);
}
}
}
}
int augment()
{
int x = t, a = INF;
while(x != s)
{
Edge& e = EG[p[x]];
a = min(a, e.cap-e.flow);
x = EG[p[x]].from;
}
x = t;
while(x != s)
{
EG[p[x]].flow += a;
EG[p[x]^1].flow -= a;
x = EG[p[x]].from;
}
return a;
}
void ISAP()
{
ans =0;
bfs();
memset(num, 0, sizeof(num));
for(int i = 0; i < n; i++)
num[d[i]]++;
int x = s;
memset(cur, 0, sizeof(cur));
while(d[s] < n)
{
if(x == t)
{
ans += augment();
x = s;
}
bool flag = false;
for(int i = cur[x]; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(e.cap > e.flow && d[x] == d[e.to]+1)
{
flag = true;
p[e.to] = G[x][i];
cur[x] = i;
x = e.to;
break;
}
}
if(!flag)
{
int m = n-1;
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(e.cap > e.flow)
m = min(m, d[e.to]);
}
if(--num[d[x]] == 0) break;
num[d[x] = m+1]++;
cur[x] = 0;
if(x != s)
x = EG[p[x]].from;
}
}
}
int main()
{
//freopen("hdu_1532.txt", "r", stdin);
int m, u, v, c;
while(~scanf("%d%d", &m, &n))
{
while(m--)
{
scanf("%d%d%d", &u, &v, &c);
addEdge(u-1, v-1, c);
}
s = 0, t = n-1;
ISAP();
printf("%d\n", ans);
EG.clear();
for(int i = 0; i < n; ++i)
G[i].clear();
}
return 0;
}