求割边最小的最小割
Harry Potter notices some Death Eaters try to slip into Castle. The Death Eaters hide in the most depths of Forbidden Forest. Harry need stop them as soon as.
The Forbidden Forest is mysterious. It consists of N nodes numbered from 0 to N-1. All of Death Eaters stay in the node numbered 0. The position of Castle is node n-1. The nodes connected by some roads. Harry need block some roads by magic and he want to minimize the cost. But it’s not enough, Harry want to know how many roads are blocked at least.
The Forbidden Forest is mysterious. It consists of N nodes numbered from 0 to N-1. All of Death Eaters stay in the node numbered 0. The position of Castle is node n-1. The nodes connected by some roads. Harry need block some roads by magic and he want to minimize the cost. But it’s not enough, Harry want to know how many roads are blocked at least.
Input
Input consists of several test cases.
The first line is number of test case.
Each test case, the first line contains two integers n, m, which means the number of nodes and edges of the graph. Each node is numbered 0 to n-1.
Following m lines contains information about edges. Each line has four integers u, v, c, d. The first two integers mean two endpoints of the edges. The third one is cost of block the edge. The fourth one means directed (d = 0) or undirected (d = 1).
Technical Specification
1. 2 <= n <= 1000
2. 0 <= m <= 100000
3. 0 <= u, v <= n-1
4. 0 < c <= 1000000
5. 0 <= d <= 1
The first line is number of test case.
Each test case, the first line contains two integers n, m, which means the number of nodes and edges of the graph. Each node is numbered 0 to n-1.
Following m lines contains information about edges. Each line has four integers u, v, c, d. The first two integers mean two endpoints of the edges. The third one is cost of block the edge. The fourth one means directed (d = 0) or undirected (d = 1).
Technical Specification
1. 2 <= n <= 1000
2. 0 <= m <= 100000
3. 0 <= u, v <= n-1
4. 0 < c <= 1000000
5. 0 <= d <= 1
Output
For each test case:
Output the case number and the answer of how many roads are blocked at least.
Output the case number and the answer of how many roads are blocked at least.
Sample Input
3
4 5
0 1 3 0
0 2 1 0
1 2 1 1
1 3 1 1
2 3 3 1
6 7
0 1 1 0
0 2 1 0
0 3 1 0
1 4 1 0
2 4 1 0
3 5 1 0
4 5 2 0
3 6
0 1 1 0
0 1 2 0
1 1 1 1
1 2 1 0
1 2 1 0
2 1 1 1
Sample Output
Case 1: 3
Case 2: 2
Case 3: 2
第一种：
建边的时候每条边权 w = w * (E + 1) + 1;
这样得到最大流 maxflow / (E + 1) ，最少割边数 maxflow % (E + 1)
道理很简单，如果原先两类割边都是最小割，那么求出的最大流相等
但边权变换后只有边数小的才是最小割了
乘（E+1）是为了保证边数叠加后依然是余数，不至于影响求最小割的结果
第二种：
建图，得到最大流后，图中边若满流，说明该边是最小割上的边
再建图，原则：满流的边改为容量为 1 的边，未满流的边改为容量 INF 的边，然后最大流即答案
程序：
#include"stdio.h"
#include"string.h"
#define M 100009
#define inf 999999999
int min(int a,int b)
{
return a<b?a:b;
}
struct st
{
int u,v,w,next;
}edge[500009];
int head[M],work[M],q[M],dis[M],t;
void init()
{
t=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
edge[t].u=u;
edge[t].v=v;
edge[t].w=w;
edge[t].next=head[u];
head[u]=t++;
edge[t].u=v;
edge[t].v=u;
edge[t].w=0;
edge[t].next=head[v];
head[v]=t++;
}
int bfs(int S,int T)
{
int rear=0;
memset(dis,-1,sizeof(dis));
q[rear++]=S;
dis[S]=0;
for(int i=0;i<rear;i++)
{
for(int j=head[q[i]];j!=-1;j=edge[j].next)
{
int v=edge[j].v;
if(edge[j].w&&dis[v]==-1)
{
dis[v]=dis[q[i]]+1;
q[rear++]=v;
if(v==T)
return 1;
}
}
}
return 0;
}
int dfs(int cur,int a,int T)
{
if(cur==T)
return a;
for(int &i=work[cur];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(edge[i].w&&dis[v]==dis[cur]+1)
{
int tt=dfs(v,min(a,edge[i].w),T);
if(tt)
{
edge[i].w-=tt;
edge[i^1].w+=tt;
return tt;
}
}
}
return 0;
}
int Dinic(int S,int T)
{
int ans=0;
while(bfs(S,T))
{
memcpy(work,head,sizeof(head));
while(int tt=dfs(S,inf,T))
ans+=tt;
}
return ans;
}
int main()
{
int T,i;
int kk=1;
scanf("%d",&T);
while(T--)
{
int n,m;
scanf("%d%d",&n,&m);
init();
while(m--)
{
int a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
add(a,b,c);
if(d==1)
add(b,a,c);
}
Dinic(0,n-1);
for(i=0;i<t;i+=2)
{
if(edge[i].w==0)
{
edge[i].w=1;
edge[i^1].w=0;
}
else
{
edge[i].w=inf;
edge[i^1].w=0;
}
}
int sum=Dinic(0,n-1);
printf("Case %d: %d\n",kk++,sum);
}
}