Drainage Ditches
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15980 Accepted Submission(s):
7594
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.
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
Recommend
正式进军省选,首先攻克网络流大关,此题是网络流之最大流问题
http://wenku.baidu.com/link?url=VTl4q3Yy3D58pfJ7IRxrlWI5Y09FSdCt6uZ8uIHUlPH2csq5WuHbBX5kPXmyFhkqT-FXxGeNYl2nEa4W1ZpLyMvpYoaE4Of-DNyfXYQ3MU_
1 #include "bits/stdc++.h" 2 #define mem(a,b) memset(a,b,sizeof(a)) 3 using namespace std; 4 typedef long long LL; 5 const int MAX=205; 6 const int oo=233333333; 7 int n,m; 8 int g[MAX][MAX],layer[MAX]; 9 bool vis[MAX]; 10 bool countlayer(){ 11 int i,j,u,v; 12 queue <int> q; 13 mem(layer,-1);layer[1]=0,q.push(1); 14 while (!q.empty()){ 15 u=q.front();q.pop(); 16 for (i=1;i<=n;i++){ 17 if (g[u][i]>0 && layer[i]==-1){ 18 layer[i]=layer[u]+1; 19 if (i==n) 20 return true; 21 else 22 q.push(i); 23 } 24 } 25 } 26 return false; 27 } 28 int dinic(){ 29 int i,j,ans(0),vs,ve; 30 int nmc,nmc_v; 31 deque <int> q; 32 while (countlayer()){ 33 q.push_back(1); 34 mem(vis,0),vis[1]=1; 35 while (!q.empty()){ 36 int nd=q.back(); 37 if (nd==n){ 38 nmc=oo; 39 for (i=1;i<q.size();i++){ 40 vs=q[i-1],ve=q[i]; 41 if (g[vs][ve]>0){ 42 if (nmc>g[vs][ve]){ 43 nmc=g[vs][ve]; 44 nmc_v=vs; 45 } 46 } 47 } 48 ans+=nmc; 49 for (i=1;i<q.size();i++){ 50 vs=q[i-1],ve=q[i]; 51 g[vs][ve]-=nmc; 52 g[ve][vs]+=nmc; 53 } 54 while (!q.empty() && q.back()!=nmc_v){ 55 vis[q.back()]=0; 56 q.pop_back(); 57 } 58 } 59 else{ 60 for (i=1;i<=n;i++){ 61 if (g[nd][i]>0 && layer[i]==layer[nd]+1 && !vis[i]){ 62 q.push_back(i); 63 vis[i]=1; 64 break; 65 } 66 } 67 if (i==n+1) 68 q.pop_back(); 69 } 70 } 71 } 72 return ans; 73 } 74 int main(){ 75 freopen ("dinic.in","r",stdin); 76 freopen ("dinic.out","w",stdout); 77 int i,j; 78 int u,v,w; 79 while (~scanf("%d%d",&m,&n)){ 80 mem(g,0); 81 for (i=1;i<=m;i++){ 82 scanf("%d%d%d",&u,&v,&w); 83 g[u][v]+=w; 84 } 85 printf("%d\n",dinic()); 86 } 87 return 0; 88 }
未来是什么样,未来会发生什么,谁也不知道。
但是我知道,
起码从今天开始努力,
肯定比从明天开始努力,
要快一天实现梦想。
千里之行,始于足下! ——《那年那兔那些事儿》