网络流几题

只说建图  代码用的模板都一样，所以只发一份就好了

POJ  1149   PIGS ： 这题建图还真有点巧妙，首先建立超级源点和汇点，源点和每个猪圈的第一个顾客连边，容量为猪圈中的猪个数，如果一个人是多个猪圈的第一个顾客，那就把这些值加起来，再连边，当然用邻接表的话这就无所谓了。 刚开始我想的是把源点和每个猪圈连边来着，然后猪圈再和每个第一位顾客连边，后来一想，没必要，而且猪圈的个数又比较多，只留顾客的话会极大的降低了图的顶点数。

然后如果顾客j在顾客i后边打开了某个猪圈，则加边i->j，容量为无穷，因为迈克可以根据顾客j的需求来将别的猪圈中的猪调整过来，所以设定为无穷大。

最后将每个顾客和汇点连边，容量为希望买的猪的数目

下面是此题的代码，我的模板要求编号是从1开始。

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <map>#include <set>#define MAXN 2222#define MAXM 222222#define INF 1000000000using namespace std;struct node{    int ver;    // vertex    int cap;    // capacity    int flow;   // current flow in this arc    int next, rev;}edge[MAXM];int dist[MAXN], numbs[MAXN], src, des, n;int head[MAXN], e;void add(int x, int y, int c){       //e记录边的总数    edge[e].ver = y;    edge[e].cap = c;    edge[e].flow = 0;    edge[e].rev = e + 1;        //反向边在edge中的下标位置    edge[e].next = head[x];   //记录以x为起点的上一条边在edge中的下标位置    head[x] = e++;           //以x为起点的边的位置    //反向边    edge[e].ver = x;    edge[e].cap = 0;  //反向边的初始网络流为0    edge[e].flow = 0;    edge[e].rev = e - 1;    edge[e].next = head[y];    head[y] = e++;}void rev_BFS(){    int Q[MAXN], qhead = 0, qtail = 0;    for(int i = 1; i <= n; ++i)    {        dist[i] = MAXN;        numbs[i] = 0;    }    Q[qtail++] = des;    dist[des] = 0;    numbs[0] = 1;    while(qhead != qtail)    {        int v = Q[qhead++];        for(int i = head[v]; i != -1; i = edge[i].next)        {            if(edge[edge[i].rev].cap == 0 || dist[edge[i].ver] < MAXN)continue;            dist[edge[i].ver] = dist[v] + 1;            ++numbs[dist[edge[i].ver]];            Q[qtail++] = edge[i].ver;        }    }}void init(){    e = 0;    memset(head, -1, sizeof(head));}int maxflow(){    int u, totalflow = 0;    int Curhead[MAXN], revpath[MAXN];    for(int i = 1; i <= n; ++i)Curhead[i] = head[i];    u = src;    while(dist[src] < n)    {        if(u == des)     // find an augmenting path        {            int augflow = INF;            for(int i = src; i != des; i = edge[Curhead[i]].ver)                augflow = min(augflow, edge[Curhead[i]].cap);            for(int i = src; i != des; i = edge[Curhead[i]].ver)            {                edge[Curhead[i]].cap -= augflow;                edge[edge[Curhead[i]].rev].cap += augflow;                edge[Curhead[i]].flow += augflow;                edge[edge[Curhead[i]].rev].flow -= augflow;            }            totalflow += augflow;            u = src;        }        int i;        for(i = Curhead[u]; i != -1; i = edge[i].next)            if(edge[i].cap > 0 && dist[u] == dist[edge[i].ver] + 1)break;        if(i != -1)     // find an admissible arc, then Advance        {            Curhead[u] = i;            revpath[edge[i].ver] = edge[i].rev;            u = edge[i].ver;        }        else        // no admissible arc, then relabel this vertex        {            if(0 == (--numbs[dist[u]]))break;    // GAP cut, Important!            Curhead[u] = head[u];            int mindist = n;            for(int j = head[u]; j != -1; j = edge[j].next)                if(edge[j].cap > 0)mindist = min(mindist, dist[edge[j].ver]);            dist[u] = mindist + 1;            ++numbs[dist[u]];            if(u != src)                u = edge[revpath[u]].ver;    // Backtrack        }    }    return totalflow;}int house[MAXN];int last[MAXN];int main(){    int m, t, k, need;    init();    scanf("%d%d", &m, &n);    src = 1;    des = n + 2;    for(int i = 1; i <= m; i++) scanf("%d", &house[i]);    for(int i = 1; i <= n; i++)    {        scanf("%d", &t);        for(int j = 0; j < t; j++)        {            scanf("%d", &k);            if(last[k] == 0) add(src, i + 1, house[k]);            else add(last[k] + 1, i + 1, INF);            last[k] = i;        }        scanf("%d", &need);        add(i + 1, des, need);    }    n = n + 2;    rev_BFS();    printf("%d\n", maxflow());    return 0;}

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