已知了最小费用流和最小割的性质，写写最大流的各种变体。 READ MORE →
图论：最小费用流与Bellman-Ford算法
最小传输费用问题
HDOJ5302：Connect the Graph——无向图构造
Problem Description
Once there was a special graph. This graph had n vertices and some edges. Each edge was either white or black. There was no edge connecting one vertex and the vertex itself. There was no two edges connecting the same pair of vertices. It is special because the each vertex is connected to at most two black edges and at most two white edges.
One day, the demon broke this graph by copying all the vertices and in one copy of the graph, the demon only keeps all the black edges, and in the other copy of the graph, the demon keeps all the white edges. Now people only knows there are w0 vertices which are connected with no white edges, w1 vertices which are connected with 1 white edges, w2 vertices which are connected with 2 white edges, b0 vertices which are connected with no black edges, b1 vertices which are connected with 1 black edges and b2 vertices which are connected with 2 black edges.
The precious graph should be fixed to guide people, so some people started to fix it. If multiple initial states satisfy the restriction described above, print any of them.
Input
The first line of the input is a single integer T (T≤700), indicating the number of testcases.
Each of the following T lines contains w0,w1,w2,b0,b1,b2. It is guaranteed that 1≤w0,w1,w2,b0,b1,b2≤2000 and b0+b1+b2=w0+w1+w2.
It is also guaranteed that the sum of all the numbers in the input file is less than 300000.
Output
For each testcase, if there is no available solution, print −1. Otherwise, print m in the first line, indicating the total number of edges. Each of the next m lines contains three integers x,y,t, which means there is an edge colored t connecting vertices x and y. t=0 means this edge white, and t=1 means this edge is black. Please be aware that this graph has no self-loop and no multiple edges. Please make sure that 1≤x,y≤b0+b1+b2.
Sample Input
2
1 1 1 1 1 1
1 2 2 1 2 2
Sample Output
-1
6
1 5 0
4 5 0
2 4 0
1 4 1
1 3 1
2 3 1
题目大意是构造一个无向图。先考虑白色边，构造常链连接1和2即可。注意排重。
参考代码：
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#include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> using namespace std; #define sf scanf int T; int a[5],b[5]; int d[1000005]; int main() { sf("%d",&T); while(T--) { for(int i = 0;i<3;i++)sf("%d",a+i); for(int i = 0;i<3;i++)sf("%d",b+i); int sum = 0; for(int i = 0;i<3;i++)sum+=a[i]; if( (a[1]&1) || (b[1]&1) ) { puts("-1"); continue; } if(sum==4) { puts("4\n1 2 0\n1 3 0\n2 3 1\n3 4 1"); continue; } printf("%d\n",a[1]/2+a[2]+b[1]/2+b[2]); int t = 1; while(a[2]!=-1){printf("%d %d 0\n",t,t+1);t++;a[2]--;}t++; while(a[1]!=2){printf("%d %d 0\n",t,t+1);t+=2;a[1]-=2;} int tt = 0; for(int i = 1;i<=sum;i+=2) d[tt++] = i; for(int i = 2;i<=sum;i+=2)d[tt++] = i; t = 0; while(b[2]!=-1){printf("%d %d 1\n",min(d[t],d[t+1]),max(d[t],d[t+1]));t++;b[2]--;}t++; while(b[1]!=2){printf("%d %d 1\n",min(d[t],d[t+1]),max(d[t],d[t+1]));t+=2;b[1]-=2;} } } |
HDOJ1733：Escape——网络流分层图
该来的总是会来的，写一道最头痛的网络流问题。
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