NOIP2008 传纸条 费用流建模
把问题抽象成图论问题,数学模型是求从S到T的两条不相交的路径,使得路径上点的权值之和最大。
费用流建模,首先拆点,把顶点i拆成i.a和i.b,i.a 与i.b之间连接一条费用为“好心值”,容量为1的有向边,特殊地,左上角和右下角两个节点拆分后点内边容量设为2,因为我们要找两条不相交路径。i右边和下边的节点j,连接一条(i.b,j.a)费用为0,容量为1的有向边。
求最大费用最大流即可,费用流值就是要求的结果。比动态规划运行得快,空间占用少。
/*
* Problem: NOIP2008 message
* Author: Guo Jiabao
* Time: 2009.6.24 11:52
* State: Solved
* Memo: 最小费用最大流
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
using namespace std;
const int MAXL=51,MAXN=MAXL*MAXL*2,MAXM=MAXN*3,INF=~0U>>1;
struct Queue
{
int Q[MAXN],head,tail,size;
bool inq[MAXN];
Queue()
{
memset(inq,0,sizeof(inq));
head=size =0;
tail=-1;
}
void ins(int p)
{
inq[p]=true;
if (++tail >= MAXN) tail = 0;
Q[tail] = p;
size ++;
}
int pop()
{
int p=Q[head];
if (++head >= MAXN) head = 0;
inq[p]=false;
size --;
return p;
}
}Q;
struct edge
{
edge *next,*op;
int t,c,v;
}ES[MAXM],*V[MAXN],*fe[MAXN];
int N,M,EC,S,T,CostFlow;
int dist[MAXN],ft[MAXN];
inline void addedge(int a,int b,int v)
{
ES[++EC].next = V[a]; V[a]=ES+EC; V[a]->t = b; V[a]->c=1; V[a]->v = v;
ES[++EC].next = V[b]; V[b]=ES+EC; V[b]->t = a; V[b]->c=0; V[b]->v = -v;
V[a]->op = V[b]; V[b]->op = V[a];
}
void init()
{
int i,j,a,b,c;
freopen("message.in","r",stdin);
freopen("message.out","w",stdout);
scanf("%d%d",&N,&M);
for (i=1;i<=N;i++)
{
for (j=1;j<=M;j++)
{
a=(i-1)*M+j;b=a+a;a=b-1;
scanf("%d",&c);
addedge(a,b,c);
}
}
ES[EC-1].c=ES[1].c=2;
for (i=1;i<=N;i++)
{
for (j=1;j<=M;j++)
{
a=((i-1)*M+j)*2;
if (j+1<=M)
{
b = ((i-1)*M+j+1)*2 - 1;
addedge(a,b,0);
}
if (i+1<=N)
{
b = (i*M+j)*2 - 1;
addedge(a,b,0);
}
}
}
S=1; T = N * M * 2;
}
bool spfa()
{
int i,j;
for (i=S;i<=T;i++)
dist[i]=-INF;
dist[S]=0;
Q.ins(S);
while (Q.size)
{
i= Q.pop();
for (edge *e=V[i];e;e=e->next)
{
j=e->t;
if (e->c && dist[i] + e->v > dist[j])
{
dist[j] = dist[i] + e->v;
ft[j] = i;
fe[j] = e;
if (!Q.inq[j])
Q.ins(j);
}
}
}
return dist[T]!=-INF;
}
void aug()
{
int i,delta=INF;
for (i=T;i!=S;i=ft[i])
if (fe[i]->c < delta)
delta = fe[i]->c;
for (i=T;i!=S;i=ft[i])
{
fe[i]->c -= delta;
fe[i]->op->c +=delta;
CostFlow += fe[i]->v * delta;
}
}
void solve()
{
while (spfa())
aug();
}
int main()
{
init();
solve();
printf("%d\n",CostFlow);
return 0;
}
上次修改时间 2017-02-03