[usaco]4.2.1 最大流问题Drainage Ditches

Drainage Ditches
Hal Burch
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. Note however, that there can be more than one ditch between two intersections.

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.

PROGRAM NAME: ditch
INPUT FORMAT
Line 1:  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. 
Line 2..N+1:  Each of 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. 

SAMPLE INPUT (file ditch.in)
5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

OUTPUT FORMAT
One line with a single integer, the maximum rate at which water may emptied from the pond.

SAMPLE OUTPUT (file ditch.out)
50

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最大流问题，按照常规算法解题即可
算法实录大致是这样的：
1：找到一个最大流路径，
2：求出其中最小的边值w；
3：把流路径上的每一条边减去w，并反向增加一条w的边。
4：重复1-3，直到找不到这样的一条路径为止。

找最大流路径的算法使用变形的dijeskal算法。

/*ID:yunleis2PROG:ditchLANG:C++*/#include<iostream>#include<fstream>using namespace std;const int MAXN=201;const int MAXM=201;int N;int M;int metri[MAXM][MAXM];int prevs[MAXM];int length[MAXM];bool visited[MAXM];int main(){fstream fin("ditch.in",ios::in);fin>>N>>M;for(int i=0;i<N;i++){int a,b,c;fin>>a>>b;fin>>c;metri[a][b]+=c;}int total=0;int flag=true;while(flag){//disjeskal to find a max flow;for(int i=1;i<=M;i++){length[i]=metri[1][i];//-metri[i][1];prevs[i]=1;visited[i]=false;}prevs[1]=0;visited[1]=true;while(true){int ptr=1;int maxflow=0;for(int i=1;i<=M;i++){if(!visited[i]&&length[i]>maxflow){maxflow=length[i];ptr=i;} }if(maxflow<=0){flag=false;break;}visited[ptr]=true;if(ptr==M){int ptr1=ptr;int minflow=1000000;while(ptr1!=1){if((metri[prevs[ptr1]][ptr1]/*-metri[ptr1][prevs[ptr1]]*/)<minflow)minflow=metri[prevs[ptr1]][ptr1];//-metri[ptr1][prevs[ptr1]];ptr1=prevs[ptr1];}ptr1=ptr;while(ptr1!=1){metri[prevs[ptr1]][ptr1]-=minflow;metri[ptr1][prevs[ptr1]]+=minflow;ptr1=prevs[ptr1];} total+=minflow;break;}for(int i=1;i<=M;i++){if(!visited[i]&&(metri[ptr][i]/*-metri[i][ptr]*/)>length[i]){length[i]=metri[ptr][i];//-metri[i][ptr];prevs[i]=ptr;}}}}fstream fout("ditch.out",ios::out);fout<<total<<endl;//system("pause");} 

USER: Ma yunlei [yunleis2]TASK: ditchLANG: C++
Compiling...Compile: OK
Executing...   Test 1: TEST OK [0.000 secs, 3184 KB]   Test 2: TEST OK [0.000 secs, 3184 KB]   Test 3: TEST OK [0.000 secs, 3184 KB]   Test 4: TEST OK [0.000 secs, 3184 KB]   Test 5: TEST OK [0.000 secs, 3184 KB]   Test 6: TEST OK [0.000 secs, 3184 KB]   Test 7: TEST OK [0.000 secs, 3184 KB]   Test 8: TEST OK [0.000 secs, 3184 KB]   Test 9: TEST OK [0.000 secs, 3184 KB]   Test 10: TEST OK [0.000 secs, 3184 KB]   Test 11: TEST OK [0.000 secs, 3184 KB]   Test 12: TEST OK [0.000 secs, 3184 KB]
All tests OK.YOUR PROGRAM ('ditch') WORKED FIRST TIME!  That's fantastic-- and a rare thing.  Please accept these special automatedcongratulations.