pid=1532">Drainage Ditches
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8574 Accepted Submission(s): 3991
Problem Description
Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's
clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
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.
clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
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.
Input
The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection
1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to
Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to
Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
Output
For each case, output a single integer, the maximum rate at which water may emptied from the pond.
Sample Input
5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10
Sample Output
50
题目大意：
就是因为下大雨的时候约翰的农场就会被雨水给淹没。无奈下约翰不得不修建水沟，并且是网络水沟，并且聪明的约翰还控制了水的流速，本题就是让你求出最大流速，无疑要运用到求最大流了。
题中m为水沟数。n为水沟的顶点，接下来Si,Ei,Ci各自是水沟的起点，终点以及其容量。求源点1到终点n的最大流速。
注意重边
</pre></p><pre name="code" class="cpp">
EdmondsKarp算法写的：
邻接矩阵：
</pre><pre name="code" class="cpp">
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f using namespace std; const int M = 1000 + 50;
int n, m;
int r[M][M];
int pre[M];// 记录结点i的前向结点为pre[i]
bool vist[M];// 记录结点i是否已訪问 bool BFS(int s, int t) //推断是否存在增广路
{
queue<int>que;
memset(pre, 0, sizeof(pre));
memset(vist, false, sizeof(vist));
pre[s] = s;
vist[s] = true;
que.push(s);
int p;
while( !que.empty() )
{
p = que.front();
que.pop();
for(int i=1; i<=n; i++)
{
if(r[p][i]>0 && !vist[i])
{
pre[i]=p;
vist[i]=true;
if( i==t )
return true;
que.push(i);
}
}
}
return false;
} int EK(int s, int t)
{
int maxflow = 0;
while( BFS(s, t) )
{
int d = INF;
// 若有增广路径，则找出最小的delta
for(int i=t; i!=s; i=pre[i])
d = min(d, r[ pre[i] ][i]);
// 这里是反向边
for(int i=t; i!=s; i=pre[i])
{
r[ pre[i] ][i] -= d;//方向边
r[i][ pre[i] ] += d;//方向边
}
maxflow += d;
}
return maxflow;
} int main()
{
while(cin>>m>>n)
{
memset(r, 0, sizeof(r));
for(int i=0; i<m; i++)
{
int from, to, rap;
scanf("%d%d%d", &from, &to, &rap);
r[from][to] += rap;
}
cout<<EK(1, n)<<endl;
}
return 0;
}
邻接表（紫书上的模板）：
#include<cstdio> #include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue> using namespace std; #define INF 0x3f3f3f3f;
const int MAXN = 1000 + 50; struct Edge
{
int from, to, cap, flow;
Edge (int u, int v, int c, int f):from(u), to(v), cap(c), flow(f) {}
}; struct EdmondsKarp
{
int n, m;
vector<Edge> edges;
vector<int> G[MAXN];
int a[MAXN];
int p[MAXN]; void init(int n)
{
for(int i=0; i<n; i++)
G[i].clear();
edges.clear();
} void AddEdge(int from, int to, int cap)
{
edges.push_back( Edge(from, to, cap, 0) );
edges.push_back( Edge(to, from, 0, 0) );
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
} int Maxflow(int s, int t)
{
int flow = 0;
for( ; ; )
{
memset(a, 0, sizeof(a));
queue<int> Q;
Q.push(s);
a[s]=INF;
while( !Q.empty() )
{
int x = Q.front();
Q.pop();
for(int i=0; i<G[x].size(); i++)
{
Edge& e = edges[ G[x][i] ];
if( !a[e.to] && e.cap > e.flow )
{
p[e.to] = G[x][i];
a[e.to] = min(a[x], e.cap-e.flow);
Q.push(e.to);
}
}
if( a[t] ) break;
}
if( ! a[t] ) break;
for(int u=t; u!=s; u=edges[ p[u] ].from )
{
edges[ p[u] ].flow += a[t];
edges[ p[u]^1 ].flow -= a[t];
}
flow += a[t];
}
return flow;
} }; int main()
{
EdmondsKarp T;
int n, m;
while(scanf("%d%d", &m, &n) !=EOF)
{
T.init(n+1);
for(int i=0; i<m; i++)
{
int a1, a2, a3;
scanf("%d%d%d", &a1, &a2, &a3);
T.AddEdge(a1, a2, a3);
}
printf("%d\n", T.Maxflow(1, n));
} return 0;
}
Dinic算法：
#include <cstdio>
#include <cstring>
#include <queue>
#define MAXN 205
#define INF 1000000000
using namespace std;
struct Edge {
int from, to, cap, flow;
}; struct Dinic {
int n, m, s, t;
vector<Edge> edges; //边表.edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表。G[i][j]表示结点i的第j条边在e数组中的序号
bool vis[MAXN]; //BFS使用
int d[MAXN]; //从起点到i的距离
int cur[MAXN]; //当前弧指针 void ClearAll(int n) {
for (int i = 0; i < n; i++) G[i].clear();
edges.clear();
} void AddEdge(int from, int to, int cap) {
edges.push_back((Edge) {from, to, cap, 0});
edges.push_back((Edge) {to, from, 0, 0});
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
} bool BFS() {//使用BFS计算出每个点在残量网络中到t的最短距离d.
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(s);
vis[s] = 1;
d[s] = 0;
while (!Q.empty()) {
int x = Q.front(); Q.pop();
for (int i = 0; i < G[x].size(); i++) {
Edge& e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow) { //仅仅考虑残量网络中的弧
vis[e.to] = 1;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
} int DFS(int x, int a) {//使用DFS从S出发，沿着d值严格递减的顺序进行多路增广。
if (x == t || a == 0) return a;
int flow = 0, f;
for (int& i = cur[x]; i < G[x].size(); i++) {
Edge& e = edges[G[x][i]];
if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[G[x][i] ^ 1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
} int Maxflow(int s, int t) {
this->s = s; this->t = t;
int flow = 0;
while (BFS()) {
memset(cur, 0, sizeof(cur));
flow += DFS(s, INF);
}
return flow;
} };
Dinic g;
int main()
{
int n, m, i, a, b, c;
while (~scanf("%d%d", &m, &n)) {
g.ClearAll(n + 1);
for (i = 0; i < m; i++) {
scanf("%d%d%d", &a, &b, &c);
g.AddEdge(a, b, c);
}
int flow = g.Maxflow(1, n);
printf("%d\n", flow);
}
return 0;
}