POJ 3436 ACM Computer Factory

题意：不想描述了。。直接从这个博客抄了：http://www.cnblogs.com/rainydays/archive/2011/07/01/2095634.html

有一些机器用来组装电脑，每台机器对输入机器的电脑有要求，符合要求的电脑被送入机器后会输出一台规定配件情况的电脑。而且分别告知每台机器在单位时间内处理电脑的台数。将这些机器连成生产线，问单位时间内出产的具有所有配件的电脑最多有多少台。

其实就是个拆点的最大流，然后要求输出所有边的流量有变化的正向边。当然也可以不拆点做。WA了一次，题目有个小坑，就是如果所有输入都是2，那么它还是可以跟源点相连的。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 60;
const int maxm = 100000;
const int INF = 1<<30;
int u[maxm],v[maxm],next[maxm],w[maxm],p[maxm],res[maxm][3];
int first[maxn],d[maxn],work[maxn],q[maxn];
int e,S,T,P,N;
int in[60][15],out[60][15],Q[60];

void init(){
    e = 0;
    memset(first,-1,sizeof(first));
}

void add_edge(int a,int b,int c){
    //printf("add:%d to %d,cap = %d
",a,b,c);
    u[e] = a;v[e] = b;next[e] = first[a];w[e] = c;first[a] = e++;
    u[e] = b;v[e] = a;next[e] = first[b];w[e] = 0;first[b] = e++;
}

int bfs(){
    int rear = 0;
    memset(d,-1,sizeof(d));
    q[rear++] = S;d[S] = 0;
    for(int i = 0;i < rear;i++){
        for(int j = first[q[i]];j != -1;j = next[j])
            if(w[j] && d[v[j]] == -1){
                d[v[j]] = d[q[i]] + 1;
                q[rear++] = v[j];
                if(v[j] == T)   return 1;
            }
    }
    return 0;
}

int dfs(int cur,int a){
    if(cur == T)    return a;
    for(int &i = work[cur];i != -1;i = next[i]){
        if(w[i] && d[v[i]] == d[cur] + 1)
            if(int t = dfs(v[i],min(a,w[i]))){
                w[i] -= t;w[i^1] += t;
                return t;
            }
    }
    return 0;
}

int dinic(){
    int ans = 0;
    while(bfs()){
        memcpy(work,first,sizeof(first));
        while(int t = dfs(S,INF))   ans += t;
    }
    return ans;
}

bool judge(int a,int b){
    for(int i = 1;i <= P;i++){
        if(out[a][i] == 0 && in[b][i] == 1) return false;
        if(out[a][i] == 1 && in[b][i] == 0) return false;
    }
    return true;
}

int main()
{
    while(scanf("%d%d",&P,&N) == 2){
        init();
        S = 0,T = N+1;
        for(int i = 1;i <= N;i++){
            scanf("%d",&Q[i]);
            int tmp = 0;
            for(int j = 1;j <= P;j++){
                scanf("%d",&in[i][j]);
                tmp += in[i][j];
            }
            if(tmp == 0 || tmp == P*2)    add_edge(S,i,Q[i]);
            tmp = 0;
            for(int j = 1;j <= P;j++){
                scanf("%d",&out[i][j]);
                tmp += out[i][j];
            }
            if(tmp == P)    add_edge(i,T,Q[i]);
        }
        int cnt = 0;
        for(int i = 1;i <= N;i++){
            for(int j = 1;j <= N;j++){
                if(i == j)  continue;
                if(judge(i,j))  p[cnt++] = e,add_edge(i,j,min(Q[i],Q[j]));
            }
        }
        printf("%d",dinic());
        int ans = 0;
        for(int i = 0;i < cnt;i++){
            if(w[p[i]^1] != 0){
                res[ans][0] = u[p[i]];
                res[ans][1] = v[p[i]];
                res[ans++][2] = w[p[i]^1];
            }
        }
        printf(" %d
",ans);
        for(int i = 0;i < ans;i++){
            printf("%d %d %d
",res[i][0],res[i][1],res[i][2]);
        }
    }
    return 0;
}