网络流-最大流:两枚[poj1459&poj3436]

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说说建图吧…

poj1459:

增加超级源点,超级汇点,跑一遍即可。

#include <cstdio>#include <cstring>#include <vector>#include <cstdlib>#include <cmath>#include <queue>#include <algorithm>using namespace std;const int MAX = 107;const int INF = 0xfffffff;struct node {    int to;    int cap;    int rev;};vector<node> G[MAX];int level[MAX];bool vis[MAX];int n, np, nc, m, S, T;inline void add_edge(int u, int v, int c) {    G[u].push_back((node){v, c, G[v].size()});    G[v].push_back((node){u, 0, G[u].size() - 1});}// from S to T, with max cap: Cbool BFS(int S, int T) {    queue<int> Q;    Q.push(S);    memset(level, -1, sizeof(level));    level[S] = 0;    while (!Q.empty()) {        int p = Q.front();        Q.pop();        for (vector<node>::iterator it = G[p].begin(); it != G[p].end(); ++it) {            if (level[it->to] < 0 && it->cap > 0) {                level[it->to] = level[p] + 1;                Q.push(it->to);                if (it->to == T) return true;            }        }    }    return false;}int DFS(int u, int v, int c) {    if (u == v) return c;    int sum = 0, tmp;    for (vector<node>::iterator it = G[u].begin(); it != G[u].end(); ++it) {        if (level[it->to] == level[u] + 1 && it->cap > 0) {            tmp = DFS(it->to, T, min(c - sum, it->cap));            sum += tmp;            it->cap -= tmp;            G[it->to][it->rev].cap += tmp;        }    }    return sum;}int dinic(int S, int T) {    int sum = 0;    while (BFS(S, T)) {        memset(vis, false, sizeof(vis));        sum += DFS(S, T, INF);    }    return sum;}int main() {    while (~scanf(" %d %d %d %d", &n, &np, &nc, &m)) {        S = n, T = n + 1;        for (int i = 0; i <= T; ++i) G[i].clear();        int a, b, c;        for (int i = 0; i < m; ++i) {            scanf(" (%d,%d)%d", &a, &b, &c);            add_edge(a, b, c);        }        for (int i = 0; i < np; ++i) {            scanf(" (%d)%d", &a, &c);            add_edge(S, a, c);        }        for (int i = 0; i < nc; ++i) {            scanf(" (%d)%d", &a, &c);            add_edge(a, T, c);        }        printf("%d\n", dinic(S, T));    }    return 0;}

poj3436 ACM Computer Factory:

题意:

好多机器生产电脑,每台机器对应的P个元器件。
输入:
0表示不能有
1表示一定要有
2表示可有可无
输出:
0表示没有
1表示
另外还有一个流量值Q,即生产能力。
举例说明,如果有3种元器件,那么机器{15, [1, 2, 0], [1, 0, 1]}表示它可同时艹(即加工,编者注)15台产品{本题是生产电脑},并且拿给它加工的产品必须有元件1(不然可能产生故障),元件2可有可无(不影响),一定不能有元件3(不然卡住了就拔不出来了);同时,经过它加工后的产品会剩下元件1和元件3,不会有元件2.
深入理解后可以建图了:
①增加超级源点,超级汇点。对每个机器判断它是否能作为源点(输入的P的元器件要求不为’1’),是否能作为汇点(输出中P种原件均为’1’,也就是产品必须是完整哒)。对应与源汇点连边,容量INF。
②对每个机器,拆点,中间流量为对应那台机器的生产能力。
③ 两两检查每对机器,我们对每对组合判断一下:
如果对某元器件,机器1输出为x,机器2要求输入为y,那么
x,y=
0,0: OK
0,1: 不行
0,2:行
1,0:不行
1,1:行
1,2:行
所以只要检查

x+y==1?不行:行

即可。

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <climits>#include <algorithm>using namespace std;const int MAX = 55;const int INF = 0xfffffff;int G[MAX << 1][MAX << 1];int in[MAX][MAX], out[MAX][MAX];int level[MAX << 1];int N, P;struct o_o {    int u;    int v;    int c;};bool BFS(int S, int T) {    queue<int> Q;    memset(level, -1, sizeof(level));    Q.push(S);    level[S] = 0;    while (!Q.empty()) {        int p = Q.front();        Q.pop();        if (p == T) return true;        //这里务必保证T的标号最大        for (int i = 0; i <= T; ++i) {            if (level[i] == -1 && G[p][i] > 0) {                level[i] = level[p] + 1;                Q.push(i);            }        }    }    return false;}int DFS(int s, int t, int flow) {    //printf("vis %d----------------\n", S);    if (s == t) return flow;    int sum = 0, tmp;    for (int i = 0; i <= t; ++i) {        if (level[i] == level[s] + 1 && G[s][i] > 0) {            tmp = DFS(i, t, min(flow - sum, G[s][i]));            sum += tmp;            //printf("(%d,%d)\n", s, sum);            G[s][i] -= tmp;            G[i][s] += tmp;        }    }    return sum;}int dinic(int S, int T) {    int sum = 0;    while (BFS(S, T)) {        //puts("---------------");        //for (int i = 0; i <= T; ++i) {        //  printf("%d ", level[i]);        //}        //puts("\n----------------");        sum += DFS(S, T, INF);        //printf("sum = %d\n", sum);        //puts("-----------------------");        //getchar();    }    return sum;}int main() {    int GT[MAX << 1][MAX << 1];    while (~scanf(" %d %d", &P, &N)) {        memset(G, 0, sizeof(G));        for (int i = 1; i <= N; ++i) {            //拆点            scanf(" %d", &G[i][i + N]);            for (int j = 1; j <= P; ++j) {                scanf(" %d", &in[i][j]);            }            for (int j = 1; j <= P; ++j) {                scanf(" %d", &out[i][j]);            }        }        int S = 0, T = N << 1 | 1;        for (int i = 1; i <= N; ++i) {            bool flag1 = true, flag2 = true;;            for (int j = 1; j <= P; ++j) {                if (in[i][j] == 1) {                    flag1 = false;                }                if (out[i][j] != 1) {                    flag2 = false;                }            }            if (flag1) G[S][i] = INF; //source            if (flag2) G[i + N][T] = INF; //slink            for (int j = 1; j <= N; ++j) {                if (i == j) continue;                flag1 = true;                for (int k = 1; k <= P; ++k) {                    if (out[i][k] + in[j][k] == 1) {                        flag1 = false;                        break;                    }                }                if (flag1) G[i + N][j] = INF;            }        }        memcpy(GT, G, sizeof(G));        int ans = dinic(S, T);        vector<o_o> V;        o_o u_u;        if (ans) {            for (int i = 1; i < T; ++i) {                for (int j = 1; j < T; ++j) {                    if (G[i][j] < GT[i][j]) {                        u_u.u = i > N ? i - N : i;                        u_u.v = j > N ? j - N : j;                        u_u.c = GT[i][j] - G[i][j];                        if (u_u.u == u_u.v) continue;                        V.push_back(u_u);                    }                }            }            printf("%d %d\n", ans, V.size());            for (vector<o_o>::iterator it = V.begin(); it != V.end(); ++it) {                printf("%d %d %d\n", it->u, it->v, it->c);            }        } else {            puts("0 0");        }    }    return 0;}
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