POJ 1149 PIGS network flow diagram + three AC methods, poj1149

POJ 1149 PIGS network flow diagram + three AC methods, poj1149

Link: POJ 1149 PIGS

PIGS Time Limit:1000 MS   Memory Limit:10000 K Total Submissions:16533   Accepted:7403 Description Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. MERs come to the farm one after another. each of them has keys to some pig-houses and wants to buy a certain number of pigs. All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold. More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pig into ss the unlocked pig-houses. An unlimited number of pigs can be placed in every pig-house. Write a program that will find the maximum number of pigs that he can operate on that day. Input The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of MERs. pig houses are numbered from 1 to M and customers are numbered from 1 to N. The next line contains M integeres, for each pig-house initial number of pig. The number of pig in each pig-house is greater or equal to 0 and less or equal to 1000. The next N lines contains records about the MERs in the following form (record about the I-th customer is written in the (I + 2)-th line ): A K1 K2... ka B It means that this customer has key to the pig-houses marked with the numbers K1, K2 ,..., KA (sorted nondecreasingly) and that he wants to buy B pigs. numbers A and B can be equal to 0. Output The first and only line of the output shoshould contain the number of sold pig. Sample Input 3 33 1 102 1 2 22 1 3 31 2 6 Sample Output 7 Source Croatia OI 2002 Final Exam-First day

Analysis:

The key to this question is how to make a diagram. After creating a graph, you can use a template to directly solve the problem. In fact, most of the network flow questions can only be illustrated. However, the diagram is a mental process.

The diagram is as follows:

(1) set a Source Vertex s and a sink vertex t,

(2) regard customers as intermediate nodes,

(3) s is connected to the first customer in each pigsty. The capacity is the number of pig in the pigsty at the beginning. If there is a duplicate edge with a node, the capacity value is merged (because the outbound traffic from the source point is the number of pigs that can be provided by all pigtails ),

(4) When customer j immediately opens a pigsty after customer I, the size of the edge <I, j> is INF, because if customer j opens a pigsty after customer I, then Mike may adjust the pig from other pig circles to this pig according to customer j's requirements, so that customer j can buy as many pig as possible,

(5) The edge size is the number of pigs the customer wants to buy.

After completing the diagram, you can solve the maximum flow. Three methods

(1) EK algorithm and shortest augmented Path Algorithm (that is, BFS is used to calculate the augmented path and then the augmented path, but BFS is required each time, which is not complex in time.

The implementation is as follows:

# Include <iostream >#include <cstdio> # include <cstring> # include <queue> # include <vector> using namespace std; # define maxn 1010 # define INF 0x3f3f3f3fint ans, s, t, n; int a [maxn], pre [maxn]; int flow [maxn] [maxn]; int cap [maxn] [maxn]; void EK () {queue <int> q; memset (flow, 0, sizeof (flow); ans = 0; while (1) {memset (a, 0, sizeof ()); a [s] = INF; q. push (s); while (! Q. empty () // bfs find the augmented path {int u = q. front (); q. pop (); for (int v = 1; v <= n; v ++) if (! A [v] & cap [u] [v]> flow [u] [v]) {pre [v] = u; q. push (v); a [v] = min (a [u], cap [u] [v]-flow [u] [v]);} if (a [t] = 0) break; for (int u = t; u! = S; u = pre [u]) // improved network stream {flow [pre [u] [u] + = a [t]; flow [u] [pre [u]-= a [t];} ans + = a [t];} int main () {// freopen ("poj_1149.txt", "r", stdin); int m, x, A, B, pig [maxn], pre [maxn]; memset (cap, 0, sizeof (cap); scanf ("% d", & m, & n); memset (pre,-1, sizeof (pre )); s = 0; t = n + 1; for (int I = 1; I <= m; I ++) scanf ("% d", & pig [I]); for (int I = 1; I <= n; I ++) {scanf ("% d", & A); for (int j = 0; j <; j ++) {scanf ("% d", & x); if (pre [x] =-1) cap [s] [I] + = pig [x]; else cap [pre [x] [I] = INF; pre [x] = I ;} scanf ("% d", & cap [I] [t]);} n + = 2; EK (); printf ("% d \ n", ans ); return 0 ;}

(2) Dinic algorithm. It is the same as EK in the SAP algorithm. It uses a hierarchy chart to find the shortest path, and then uses DFS to expand.

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <vector>using namespace std;#define maxn 1010#define INF 0x3f3f3f3fstruct Edge{    int from, to, cap;};vector<Edge> EG;vector<int> G[maxn];int n, s, t, ans, d[maxn], cur[maxn];void addEdge(int from, int to, int cap){    EG.push_back((Edge){from, to, cap});    EG.push_back((Edge){to, from, 0});    int x = EG.size();    G[from].push_back(x-2);    G[to].push_back(x-1);}bool bfs(){    memset(d, -1, sizeof(d));    queue<int> q;    q.push(s);    d[s] = 0;    while(!q.empty())    {        int x = q.front();        q.pop();        for(int i = 0; i < G[x].size(); i++)        {            Edge& e = EG[G[x][i]];            if(d[e.to] == -1 && e.cap > 0)            {                d[e.to] = d[x]+1;                q.push(e.to);            }        }    }    return (d[t]!=-1);}int dfs(int x, int a){    if(x == t || a == 0) return a;    int flow = 0, f;    for(int& i = cur[x]; i < G[x].size(); i++)    {        Edge& e = EG[G[x][i]];        if(d[x]+1 == d[e.to] && (f = dfs(e.to, min(a, e.cap))) > 0)        {            e.cap -= f;            EG[G[x][i]^1].cap += f;            flow += f;            a -= f;            if(a == 0) break;        }    }    return flow;}void Dinic(){    ans = 0;    while(bfs())    {        memset(cur, 0, sizeof(cur));        ans += dfs(s, INF);    }}int main(){    //freopen("poj_1149.txt", "r", stdin);    int m, x, A, B, pig[maxn], pre[maxn];    scanf("%d%d", &m, &n);    memset(pre, -1, sizeof(pre));    s = 0; t = n+1;    for(int i = 1; i <= m; i++)        scanf("%d", &pig[i]);    for(int i = 1; i <= n; i++) {        scanf("%d", &A);        for(int j = 0; j < A; j++) {            scanf("%d", &x);            if(pre[x] == -1)                addEdge(s, i, pig[x]);            else                addEdge(pre[x], i, INF);            pre[x] = i;        }        scanf("%d", &B);        addEdge(i, t, B);    }    n += 2;    Dinic();    printf("%d\n", ans);    return 0;}

(3) ISAP, improving the ISAP algorithm, is currently the best algorithm. It also finds the shortest path through the layered network, but the reverse BFS, and BFS only needs to run once

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <vector>using namespace std;#define maxn 1010#define INF 0x3f3f3f3fstruct Edge{    int from, to, cap, flow;};vector<Edge> EG;vector<int> G[maxn];int n, s, t, ans, d[maxn], cur[maxn], p[maxn], num[maxn];bool vis[maxn];void addEdge(int from, int to, int cap){    EG.push_back((Edge){from, to, cap, 0});    EG.push_back((Edge){to, from, 0, 0});    int x = EG.size();    G[from].push_back(x-2);    G[to].push_back(x-1);}void bfs(){    memset(vis, false, sizeof(vis));    queue<int> q;    vis[t] = true;    d[t] = 0;    q.push(t);    while(!q.empty()) {        int x = q.front();        q.pop();        for(int i = 0; i < G[x].size(); i++) {            Edge& e = EG[G[x][i]^1];            if(!vis[e.from] && e.cap > e.flow) {                vis[e.from] = true;                d[e.from] = d[x]+1;                q.push(e.from);            }        }    }}int augment(){    int x = t, a = INF;    while(x != s) {        Edge& e = EG[p[x]];        a = min(a, e.cap-e.flow);        x = EG[p[x]].from;    }    x = t;    while(x != s) {        EG[p[x]].flow += a;        EG[p[x]^1].flow -= a;        x = EG[p[x]].from;    }    return a;}void ISAP(){    ans =0;    bfs();    memset(num, 0, sizeof(num));    for(int i = 0; i < n; i++)        num[d[i]]++;    int x = s;    memset(cur, 0, sizeof(cur));    while(d[s] < n) {        if(x == t) {            ans += augment();            x = s;        }        bool flag = false;        for(int i = cur[x]; i < G[x].size(); i++) {            Edge& e = EG[G[x][i]];            if(e.cap > e.flow && d[x] == d[e.to]+1) {                flag = true;                p[e.to] = G[x][i];                cur[x] = i;                x = e.to;                break;            }        }        if(!flag) {            int m = n-1;            for(int i = 0; i < G[x].size(); i++) {                Edge& e = EG[G[x][i]];                if(e.cap > e.flow)                    m = min(m, d[e.to]);            }            if(--num[d[x]] == 0) break;            num[d[x] = m+1]++;            cur[x] = 0;            if(x != s)                x = EG[p[x]].from;        }    }}void Clear(){    EG.clear();    for(int i = 0; i < n; ++i)        G[i].clear();}int main(){    //freopen("poj_1149.txt", "r", stdin);    int m, x, A, B, pig[maxn], pre[maxn];    scanf("%d%d", &m, &n);        memset(pre, -1, sizeof(pre));        s = 0; t = n+1;        for(int i = 1; i <= m; i++)            scanf("%d", &pig[i]);        for(int i = 1; i <= n; i++)        {            scanf("%d", &A);            for(int j = 0; j < A; j++)            {                scanf("%d", &x);                if(pre[x] == -1)                    addEdge(s, i, pig[x]);                else                    addEdge(pre[x], i, INF);                pre[x] = i;            }            scanf("%d", &B);            addEdge(i, t, B);        }        n += 2;        ISAP();        printf("%d\n", ans);        Clear();    return 0;}

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