poj3436 ACM Computer Factory

首先题目不是本英语渣渣能读懂的了，看的网上的解释。

一个工厂里面有n台机器，每台机器能同时加工电脑的总数是有限的（点有容量），所以要拆点，在就是建立超级源点和汇点。

每台电脑有p个零件，但是不是按照任意顺序添加上去就是可以的，有一定添加顺序关系，同时每台机器接收的需加工的电脑是有要求的，电脑上面那些零件必须有1，那些不能有0，那些可有可无2，而加工后电脑上面的零件有为1，无为0。

两台机器可以建立关系的条件就是前一台机器生产的电脑的条件满足后面这台机器接收电脑的条件，容量为前一台机器的点容量，

和源点与汇点的链接也是同上面的判断原理一样，剩下的就是最大流了。

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2016File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtypedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;typedef pair<ii,int> iii;const double eps = 1e-10;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;int p, n;int s, t;const int maxn = 110;int can[maxn], in[maxn][maxn], out[maxn][maxn];int cap[maxn*2][maxn*2], flow[maxn*2][maxn*2], pre[maxn*2];inline void Input() {    memset(cap, 0,sizeof cap);    memset(flow, 0,sizeof flow);    for (int i = 1;i <= n;++i) {        scanf("%d",&can[i]);        cap[i][i+n] = can[i];//拆点        for (int j = 1;j <= p;++j)            scanf("%d",&in[i][j]);        for (int j = 1;j <= p;++j)            scanf("%d",&out[i][j]);    }    // for (int i = 1;i <= n;++i)    //     cout << can[i] << ' ';    // cout << endl;    s = 2*n+1,t = 2*n+2;}bool Judge1(int x) {    for (int i = 1;i <= p;++i)        if (in[x][i]==1) return false;    return true;}bool Judge2(int x) {    for (int i = 1;i <= p;++i)        if (!out[x][i]) return false;    return true;}bool Judge3(int x,int y) {    for (int i = 1;i <= p;++i)        if (out[x][i] + in[y][i] == 1) return false;    return true;}inline void GetMap() {    for (int i = 1;i <= n;++i) {        if (Judge1(i)) {            cap[s][i] = INF;            // cout << "i = " << i << endl;        }        if (Judge2(i)) {             cap[i+n][t] = can[i];            // cout << "g = " << i + n << endl;        }        for (int j = 1;j <= n;++j) {            if (i==j) continue;            if (Judge3(i, j)) {                // printf("%d->%d\n", i+n,j);                cap[i+n][j] = can[i];            }        }    }}int dis[maxn*2];bool vis[maxn*2];bool spfa() {    queue<int> que;    memset(dis, -1,sizeof dis);    dis[s] = 0;    que.push(s);    while(!que.empty()) {        int u = que.front();        // cout << u << ' ';        que.pop();        for (int v = 1;v <= t;++v) {            if (cap[u][v] - flow[u][v] > 0 && dis[v] == -1) {                dis[v] = dis[u] + 1;                que.push(v);            }        }    }    // cout << endl;    return dis[t] != -1;}int dfs(int u,int a) {    if (u == t || a == 0) return a;    int ret = 0,f;    for (int v = 1;v <= t;++v) {        if (u != v && dis[v] >= dis[u] + 1 && (f = dfs(v, min(a,cap[u][v] - flow[u][v]))) > 0) {            ret += f;            a -= f;            flow[u][v] += f;            flow[v][u] -= f;            if (a == 0) break;        }    }    return ret;}int solve() {    int ret = 0;    while(spfa()) {        // cout << "here\n";        ret += dfs(s, INF);        // break;    }    return ret;}int answer[maxn*2][3];int cnt;inline void Fock() {    cnt = 0;    for (int i = n + 1;i <= n*2;++i) {        for (int j = 1;j <= n;++j) {            if (flow[i][j] > 0) {                // printf("%d %d %d\n",i - n, j, flow[i][j]);                answer[cnt][0] = i - n;                answer[cnt][1] = j;                answer[cnt][2] = flow[i][j];                cnt++;            }        }    }}int main(){        // freopen("in.txt","r",stdin);    // freopen("out.txt","w",stdout);    while(~scanf("%d%d",&p,&n)) {        Input();        GetMap();        int ans = solve();        Fock();        printf("%d %d\n",ans, cnt);        for (int i = 0;i < cnt;++i)            printf("%d %d %d\n", answer[i][0],answer[i][1],answer[i][2]);    }    return 0;}