POJ1149 PIGS 网络最大流

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. Customers 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 pigs across 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 sell 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 customers. 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 pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000. 
The next N lines contains records about the customers 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 should contain the number of sold pigs.

Sample Input

3 33 1 102 1 2 22 1 3 31 2 6

Sample Output

7

一般增广路径方法——Ford-Fulkerson算法（标号法）

468K0MSG++1601B

#include <cstdio>#include <cstring>#define INF 0x3f3f3f3fint s, t, m, n;int c[105][105], f[105][105];int house[1005], pre[1005];int queue[105], qs, qe;int mark[105], minf[105];void init(){int num, k;    memset(pre, 0, sizeof(pre));    memset(c, 0, sizeof(c));    scanf("%d %d", &m, &n);    s = 0; t = n + 1;    for(int i = 1; i <= m; i++)    scanf("%d", house + i);    for(int i = 1; i <= n; i++){        scanf("%d", &num);        while(num--){        scanf("%d", &k);        if(!pre[k])        c[s][i] += house[k];        else        c[ pre[k] ][i] = INF;        pre[k] = i;        }        scanf("%d", &c[i][t]);    }}void ford(){int v, p;    memset(f, 0, sizeof(f));    minf[0] = INF;    while(true){        for(int i = 0; i <= n + 1; i++)        mark[i] = -2;        mark[0] = -1;                qs = 0; queue[qs] = 0; qe = 1;        while( qs != qe && mark[t] == -2){        v = queue[qs];  qs = (qs + 1) % n;        for(int i = 0; i <= t; i ++)        if( mark[i] == -2 && (p = c[v][i] - f[v][i]) ){        mark[i] = v; queue[qe] = i; qe = (qe + 1) % n;        minf[i] = minf[v] < p ? minf[v] : p;        }        }        if(mark[t] == -2) break;        for(int i = mark[t], j = t;  i != -1;  j = i,i = mark[i] ){        f[i][j] += minf[t];        f[j][i] -= f[i][j];        }           }    int ans = 0;    for(int i = 1; i <= n; i++)    ans += f[i][t];    printf("%d\n", ans);}int main(){    init();       ford();    return 0;}

最短增广路算法——SAP

396K16MSG++1673B

#include <cstdio>#include <cstring>#define inf 0x3f3f3f3fint n, m, s, t;int c[105][105], h[105], vh[105];void init(){scanf("%d %d", &m, &n);    int a[m], b[m];    memset(a, 0, sizeof(a));    memset(vh, 0, sizeof(vh));    memset(h, 0, sizeof(h));    memset(c, 0, sizeof(c));    for(int i = 0; i < m; i++)        scanf("%d", &b[i]);    for(int i = 0; i < n; i++){        int k, l, temp;        scanf("%d", &k);        for(int j = 0; j < k; j++){            scanf("%d", &temp);            if(!a[temp - 1])                c[0][i+1] += b[temp-1];            else c[ a[temp-1] ][i+1] = inf;            a[temp-1] = i + 1;        }        scanf("%d", &l);        c[i+1][n+1] = l;    }}int sap(int i, int f){    if(i == t)        return f;    int old = f, minh = n - 1;    for(int j = 0; j < n; j++){        if(c[i][j] > 0){            if(h[i] == h[j] + 1){                int d = c[i][j] < old ? c[i][j] : old;                d = sap(j, d);                c[i][j] -= d;                c[j][i] += d;                old -= d;                if(h[s] >= n)                return  f - old;                if(old == 0)                    break;            }            if(h[j] < minh)                minh = h[j];        }    }    if(f == old){        vh[ h[i] ]--;        if(vh[ h[i] ] == 0)            h[s] = n;        h[i] = minh+1;        vh[ h[i] ] ++;    }    return f - old;}int main(){init();    s = 0, t = n + 1, vh[0] = n + 2;    int ff = 0;    n += 2;    while(h[s] < n)        ff += sap(s, inf);    printf("%d\n", ff);    return 0;}

连续最短增广路算法——Dinic   

400K0MSG++2270B

#include <cstdio>#include <cstring>#define INF 0x3f3f3f3fint n, m, s, t;int house[1005], pre[1005];int c[105][105], ps[105], dep[105];void init(){int num, k;    memset(pre, 0, sizeof(pre));    memset(c, 0, sizeof(c));    scanf("%d %d", &m, &n);    s = 0; t = n + 1;    for(int i = 1; i <= m; i++)    scanf("%d", house + i);    for(int i = 1; i <= n; i++){        scanf("%d", &num);        while(num--){        scanf("%d", &k);        if(!pre[k])        c[s][i] += house[k];        else        c[ pre[k] ][i] = INF;        pre[k] = i;        }        scanf("%d", &c[i][t]);    }}bool bfs(int n,int s,int t){    int f = 0, r = 0, u, v;    memset(dep, -1, sizeof(dep));    ps[r++] = s;    dep[s] = 0; //开始计算层次    while(f < r){        u = ps[f++];        for(v = 0; v < n; v++)            if(c[u][v] && dep[v] < 0){                dep[v] = dep[u] + 1;                ps[r++] = v;            }            if(u == t) break;    }    return dep[t] < 0; //返回能否增广}int Dinic(int n,int s,int t){    int k, u, v, num, res = 0, top, tag;    while(1){        if( bfs(n, s, t) ) break; //不能增广，最终结果        top = 0;        u = s;        while(1){            if(u == t){  //一条增广                ps[top++] = t;                for(k = 0, num = INF; k < top - 1; k++)                    if(c[ ps[k] ][ ps[k+1] ] < num){                        tag = k; //标志最小弧的头                        num = c[ ps[k] ][ ps[k+1] ];                    }                for(k=0;k<top-1;k++){                    c[ps[k]][ps[k+1]] -= num;                    c[ps[k+1]][ps[k]] += num;                }                res += num;                top = tag; //从标志处继续找其他增广路                u = tag;            }            for(v = 0; v < n; v++)                 if(c[u][v] && dep[u] + 1 == dep[v]) break;  //从u出发存在弧u->v            if(v<n) {ps[top++] = u; u = v;}   //压u入栈，从v出发找            else{                if(top == 0) break;  //确定增广路找不到                dep[u] = -1;                u=ps[--top]; //把u弹出，找层次图中另一条u->v            }        }    }    return res;}int main(){init();    printf("%d\n", Dinic(n + 2, s, t));return 0;}