POJ - 3436 ACM Computer Factory ( 最大流 )

POJ - 3436

題意：... 大意就是有幾臺機器可以把半成品加工成成品或者另外的半成品，問你最終每小時得到的最多的成品。

思路：那麼由題目，這個機器出來的零件個數和另一個機器需要的零件相同時（2的位置不算），他們肯定有條線，但是這裏可以看出，對機器每小時有限制數量，這是對點有限制，而不是對邊有限制，所以需要拆點，我第一遍寫的時候沒拆，但是過了，估計數據比較弱.....

沒拆點的：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn = 100,maxe = 1e4 + 50;
const int inf = 0x3f3f3f3f;
struct ma
{
    int sum,in[12],out[12];
}data[maxn];

struct node
{
    int to,next,cap,rem,rev;
    node(){}
    node(int a,int b,int c,int d,int e)
    {to = a; next = b; cap = c; rem = d;rev = e;}
}edge[maxe << 1];
int h[maxn],cur[maxn],deg[maxn];
int out[maxe][3];
int num ,s,t,p,n;
void init()
{
    for(int i = 0;i < maxn; i++)
        h[i] = -1;
    num = 0;    s = 0, t = n+1;
}
void add(int u,int v,int cap)
{
    edge[num] = node(v,h[u],cap,cap,num+1); h[u] = num++;
    edge[num] = node(u,h[v],0,0,num-1);   h[v] = num++;
}
void pre()
{
    for(int i = 1; i <= n; i++)
    {
        int mark = 1;
        for(int j = 0; j < p; j++)
        {
            if(data[i].in[j] == 2 ) continue;
            else if(data[i].in[j] == 1) {mark = 0; break;}
        }
        if(mark) add(s,i,data[i].sum);
    }
    for(int i = 1; i <= n; i++)
    {
        int sum = 0;
        for(int j = 0; j < p; j++)
            sum += data[i].out[j];
        if(sum == p) add(i,t,data[i].sum);
    }
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= n; j++)
    {
        int mark = 1;
        for(int k = 0; k < p;k ++)
        {
            if(data[j].in[k]==2) continue;
            else if(data[j].in[k] != data[i].out[k]) {mark = 0; break;}
        }
        if(mark) add(i,j,data[i].sum);
    }
}
int bfs()
{
    for(int i = 1; i <= n+1; i++)
        deg[i] = -1;
    queue<int > q;
    q.push(s);
    deg[s] = 0;
    while(!q.empty())
    {
        int u = q.front(); q.pop();
       //cout << "u  "<< u <<endl;
        for(int i = h[u] ;~i; i = edge[i].next)
        {
            int v = edge[i].to,rem = edge[i].rem;
            if(rem && deg[v] == -1)
            {
                deg[v] = deg[u] + 1;
                q.push(v);
            }
        }
    }
    return deg[t] != -1;
}

int dfs(int u,int f)
{
    if(u == t) return f;
    for(int i = cur[u] ;~i; i = edge[i].next)
    {
        cur[u] = i;
        int v = edge[i].to,rem = edge[i].rem,rev = edge[i].rev;
        if(deg[v] == deg[u]+1 && rem)
        {
            int k = dfs(v,min(f,rem));
            if(!k) continue;
            edge[i].rem -= k;
            edge[rev].rem += k;
            return k;
        }
    }
    return 0;
}
void solve()
{
    int flow = 0;
    while(bfs())
    {
        for(int i = s; i <= t; i++)
            cur[i] = h[i];
        for(;;)
        {
            int k = dfs(s,inf);
            if(k == 0) break;
            flow += k;
        }
    }
    printf("%d ",flow);
    int onum = 0;
    for(int u = 1; u <= n; u++)
    for(int i = h[u]; ~i; i = edge[i].next)
    {
        int v = edge[i].to, rem = edge[i].rem, cap = edge[i].cap;
        if(cap && rem < cap && v <= n)
        {
            out[onum][0] = u, out[onum][1] = v, out[onum++][2] = cap-rem;
        }
    }
    printf("%d\n",onum);
    for(int i = 0; i < onum; i++)
        printf("%d %d %d\n", out[i][0] , out[i][1] , out[i][2]);
}
int main()
{
    while(~scanf("%d%d",&p,&n))
    {
        init();
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&data[i].sum);
            for(int j = 0; j < p; j++)  scanf("%d",&data[i].in[j]);
            for(int j = 0; j < p; j++)  scanf("%d",&data[i].out[j]);
        }
        pre();
        solve();
    }
    return 0;
}

拆點

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn = 200,maxe = maxn * maxn + 50;
const int inf = 0x3f3f3f3f;
struct ma
{
    int sum,in[12],out[12];
}data[maxn];

struct node
{
    int to,next,cap,rem,rev;
    node(){}
    node(int a,int b,int c,int d,int e)
    {to = a; next = b; cap = c; rem = d;rev = e;}
}edge[maxe << 1];
int h[maxn],cur[maxn],deg[maxn];
int out[maxe][3];
int num ,s,t,p,n;
void init()
{
    for(int i = 0;i < maxn; i++)
        h[i] = -1;
    num = 0;    s = 0, t = 2*n+1;
}
void add(int u,int v,int cap)
{
    edge[num] = node(v,h[u],cap,cap,num+1); h[u] = num++;
    edge[num] = node(u,h[v],0,0,num-1);   h[v] = num++;
}
void pre()
{
    for(int i = 1; i <= n; i++)
    {
        int mark = 1;
        for(int j = 0; j < p; j++)
        {
            if(data[i].in[j] == 2 ) continue;
            else if(data[i].in[j] == 1) {mark = 0; break;}
        }
        if(mark) add(s,i,inf);
    }
    for(int i = 1; i <= n; i++) add(i,i+n,data[i].sum);
    for(int i = 1; i <= n; i++)
    {
        int sum = 0;
        for(int j = 0; j < p; j++)
            sum += data[i].out[j];
        if(sum == p) add(i+n,t,inf);
    }
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= n; j++)
    {
        if(i==j)continue;
        int mark = 1;
        for(int k = 0; k < p;k ++)
        {
            if(data[j].in[k]==2) continue;
            else if(data[j].in[k] != data[i].out[k]) {mark = 0; break;}
        }
        if(mark) add(i+n,j,inf);
    }
}
int bfs()
{
    for(int i = 1; i <= t; i++)
        deg[i] = -1;
    queue<int > q;
    q.push(s);
    deg[s] = 0;
    while(!q.empty())
    {
        int u = q.front(); q.pop();
       //cout << "u  "<< u <<endl;
        for(int i = h[u] ;~i; i = edge[i].next)
        {
            int v = edge[i].to,rem = edge[i].rem;
            if(rem && deg[v] == -1)
            {
                deg[v] = deg[u] + 1;
                q.push(v);
            }
        }
    }
    return deg[t] != -1;
}

int dfs(int u,int f)
{
    if(u == t) return f;
    for(int i = cur[u] ;~i; i = edge[i].next)
    {
        cur[u] = i;
        int v = edge[i].to,rem = edge[i].rem,rev = edge[i].rev;
        if(deg[v] == deg[u]+1 && rem)
        {
            int k = dfs(v,min(f,rem));
            if(!k) continue;
            edge[i].rem -= k;
            edge[rev].rem += k;
            return k;
        }
    }
    return 0;
}
void solve()
{
    int flow = 0;
    while(bfs())
    {
        for(int i = s; i <= t; i++)
            cur[i] = h[i];
        for(;;)
        {
            int k = dfs(s,inf);
            if(k == 0) break;
            flow += k;
        }
    }
    printf("%d ",flow);
    int onum = 0;
    for(int u = n+1; u <= n*n; u++)
    for(int i = h[u]; ~i; i = edge[i].next)
    {
        int v = edge[i].to, rem = edge[i].rem, cap = edge[i].cap;
        if(cap && rem < cap && v <= n)
            out[onum][0] = u-n, out[onum][1] = v, out[onum++][2] = cap-rem;
    }
    printf("%d\n",onum);
    for(int i = 0; i < onum; i++)
        printf("%d %d %d\n", out[i][0] , out[i][1] , out[i][2]);
}
int main()
{
    while(~scanf("%d%d",&p,&n))
    {
        init();
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&data[i].sum);
            for(int j = 0; j < p; j++)  scanf("%d",&data[i].in[j]);
            for(int j = 0; j < p; j++)  scanf("%d",&data[i].out[j]);
        }
        pre();
        solve();
    }
    return 0;
}