Drainage Ditches
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 8431 Accepted
Submission(s): 3921
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.
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.
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
Source
题意：
最大流
分析：
最大流
代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <queue>
5 #include <vector>
6 using namespace std;
7
8 const int maxn = 205 << 1;
9 const int INF = 1000000000;
10
11 struct Edge
12 {
13 int from, to, cap, flow;
14 };
15
16 struct Dinic
17 {
18 int n, m, s, t;
19 vector<Edge> edges;
20 vector<int>G[maxn];
21 bool vis[maxn];
22 int d[maxn];
23 int cur[maxn];
24
25 void ClearAll(int n) {
26 for(int i = 0; i <= n; i++) {
27 G[i].clear();
28 }
29 edges.clear();
30 }
31
32 void AddEdge(int from, int to, int cap) {
33 edges.push_back((Edge){from, to, cap, 0} );
34 edges.push_back((Edge){to, from, 0, 0} );
35 m = edges.size();
36 G[from].push_back(m - 2);
37 G[to].push_back(m - 1);
38 //printf("%din end\n",m);
39 }
40
41 bool BFS()
42 {
43 memset(vis, 0, sizeof(vis) );
44 queue<int> Q;
45 Q.push(s);
46 vis[s] = 1;
47 d[s] = 0;
48 while(!Q.empty() ){
49 int x = Q.front(); Q.pop();
50 for(int i = 0; i < G[x].size(); i++) {
51 Edge& e = edges[G[x][i]];
52 if(!vis[e.to] && e.cap > e.flow) {
53 vis[e.to] = 1;
54 d[e.to] = d[x] + 1;
55 Q.push(e.to);
56 }
57 }
58 }
59 return vis[t];
60 }
61
62 int DFS(int x, int a) {
63 if(x == t || a == 0) return a;
64 int flow = 0, f;
65 for(int& i = cur[x]; i < G[x].size(); i++) {
66 Edge& e = edges[G[x][i]];
67 if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
68 e.flow += f;
69 edges[G[x][i]^1].flow -= f;
70 flow += f;
71 a -= f;
72 if(a == 0) break;
73 }
74 }
75 return flow;
76 }
77
78 int Maxflow(int s, int t) {
79 this -> s = s; this -> t = t;
80 int flow = 0;
81 while(BFS()) {
82 memset(cur, 0, sizeof(cur) );
83 flow += DFS(s, INF);
84 }
85 return flow;
86 }
87 };
88
89 int main() {
90 int n, m;
91 int a, b, c;
92 while(EOF != scanf("%d %d",&n, &m)) {
93 Dinic g;
94 for(int i = 0; i < n; i++) {
95 scanf("%d %d %d",&a, &b, &c);
96 g.AddEdge(a, b, c);
97 }
98 printf("%d\n",g.Maxflow(1, m) );
99 }
100 return 0;
101 }
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