hdu 3987 Harry Potter and the Forbidden Forest//边数最小的最小割 2种解题方法
每条边的权值改为w*(M+1)+1
最后得到的和sum
sum div (M+1) 得到的是最小割的值，用(M+1)是为了不被后面的1影响到
sum mod (M+1)得到的是最小割的边数，这个是在上面的最小割前提下求出来的（YY一下，在原题中可能存在多组割边，边数不同，但是
值相同。现在同乘以（M+1)的话，值不变。但是+1的话，就会按照边的数量得出边数最小的最小割）
#include <cstdio>
#include <cstring>
const int N=2000 + 123;
const long long M=100000*4+12345;
const long long inf=(1ll)<<60;
int head[N];
struct Edge
{
int v,next;
long long w;
} edge[M];
int cnt,n,s,t;
void addedge(int u,int v,long long w)
{
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].v=u;
edge[cnt].w=0;
edge[cnt].next=head[v];
head[v]=cnt++;
}
int pre[N],cur[N], gap[N], dis[N];
long long sap()
{
long long flow=0,aug=inf;
int u;
bool flag;
for(int i=0; i<n; i++)
{
cur[i]=head[i];
gap[i]=dis[i]=0;
}
gap[s]=n;
u=pre[s]=s;
while(dis[s]<n)
{
flag=0;
for(int &j=cur[u]; j!=-1; j=edge[j].next)
{
int v=edge[j].v;
if(edge[j].w>0&&dis[u]==dis[v]+1)
{
flag=1;
if(edge[j].w<aug) aug=edge[j].w;
pre[v]=u;
u=v;
if(u==t)
{
flow+=aug;
while(u!=s)
{
u=pre[u];
edge[cur[u]].w-=aug;
edge[cur[u]^1].w+=aug;
}
aug=inf;
}
break;
}
}
if(flag) continue;
int mindis=n;
for(int j=head[u]; j!=-1; j=edge[j].next)
{
int v=edge[j].v;
if(edge[j].w>0&&dis[v]<mindis)
{
mindis=dis[v];
cur[u]=j;
}
}
if((--gap[dis[u]])==0) break;
gap[dis[u]=mindis+1]++;
u=pre[u];
}
return flow;
}
int main()
{
int T, cas=1;
int m;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
s=0;
t=n -1;
cnt=0;
memset(head,-1,sizeof(head));
while(m--)
{
int a, b,d;
long long c;
scanf("%d%d%I64d%d", &a, &b, &c, &d);
if(d == 0)
{
addedge(a, b, M * c + 1);
}
else
{
addedge(a, b, M * c + 1);
addedge(b, a, M * c + 1);
}
}
long long ans = sap() % M;
printf("Case %d: %I64d\n", cas++, ans);
}
return 0;
}
这个方法比较传统，不多说了
#include <cstdio>
#include <cstring>
const int N=2000 + 123;
const long long M=100000*4+12345;
const long long inf=(1ll)<<60;
int head[N];
struct Edge
{
int v,next;
long long w;
} edge[M], ed[M];
struct bian
{
int u, v, d, id[2];
long long c;
}bian[M];
int cnt,n,s,t;
void addedge(int u,int v,long long w)
{
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].v=u;
edge[cnt].w=0;
edge[cnt].next=head[v];
head[v]=cnt++;
}
void addedge2(int u,int v,long long w)
{
ed[cnt].v=v;
ed[cnt].w=w;
ed[cnt].next=head[u];
head[u]=cnt++;
ed[cnt].v=u;
ed[cnt].w=0;
ed[cnt].next=head[v];
head[v]=cnt++;
}
int pre[N],cur[N], gap[N], dis[N];
long long sap(Edge * edge)
{
long long flow=0,aug=inf;
int u;
bool flag;
for(int i=0; i<n; i++)
{
cur[i]=head[i];
gap[i]=dis[i]=0;
}
gap[s]=n;
u=pre[s]=s;
while(dis[s]<n)
{
flag=0;
for(int &j=cur[u]; j!=-1; j=edge[j].next)
{
int v=edge[j].v;
if(edge[j].w>0&&dis[u]==dis[v]+1)
{
flag=1;
if(edge[j].w<aug) aug=edge[j].w;
pre[v]=u;
u=v;
if(u==t)
{
flow+=aug;
while(u!=s)
{
u=pre[u];
edge[cur[u]].w-=aug;
edge[cur[u]^1].w+=aug;
}
aug=inf;
}
break;
}
}
if(flag) continue;
int mindis=n;
for(int j=head[u]; j!=-1; j=edge[j].next)
{
int v=edge[j].v;
if(edge[j].w>0&&dis[v]<mindis)
{
mindis=dis[v];
cur[u]=j;
}
}
if((--gap[dis[u]])==0) break;
gap[dis[u]=mindis+1]++;
u=pre[u];
}
return flow;
}
int main()
{
int T, cas=1;
int m;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
s=0;
t=n -1;
cnt=0;
memset(head,-1,sizeof(head));
for(int i = 0; i < m; i++)
{
int a, b,d;
long long c;
scanf("%d%d%I64d%d", &bian[i].u, &bian[i].v, &bian[i].c, &bian[i].d);
if(bian[i].d == 0)
{
bian[i].id[0] = cnt;
addedge(bian[i].u, bian[i].v, bian[i].c);
}
else
{
bian[i].id[0] = cnt;
addedge(bian[i].u, bian[i].v, bian[i].c);
bian[i].id[1] = cnt;
addedge(bian[i].v, bian[i].u, bian[i].c);
}
}
long long ans = sap(edge) ;
//printf("Case %d: %I64d\n", cas++, ans);
cnt=0;
memset(head,-1,sizeof(head));
for(int i = 0; i < m; i++)
{
if(bian[i].d == 0)
{
if(edge[bian[i].id[0]].w == 0) //满流
addedge2(bian[i].u, bian[i].v, 1);
else
addedge2(bian[i].u, bian[i].v, inf);
}
else
{
if(edge[bian[i].id[0]].w == 0) //满流
addedge2(bian[i].u, bian[i].v, 1);
else
addedge2(bian[i].u, bian[i].v, inf);
if(edge[bian[i].id[1]].w == 0) //满流
addedge2(bian[i].v, bian[i].u, 1);
else
addedge2(bian[i].v, bian[i].u, inf);
}
}
printf("Case %d: %I64d\n", cas++, sap(ed));
}
return 0;
}