UVa 259 Software Allocation （ 最大流 ）

这道题也是，输入很坑，每组之间也空格结束，最后以EOF结束

建图其实有一种比这个好的方法，不用拆点，就是源点和computer相连，容量1， computer和program连，容量1， program和汇点连，容量为相应数字

但是一下代码，是拆点做的，把computer拆了，因为computer的容量为1，下面代码也是对照这个建图方法建的，program和源点连，容量为相应数字，和computer连，容量为1，computer容量为1， computer拆出来的点，和汇点连，容量INF

代码如下：

//by Molly//Maximum Flow#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#include <string>#include <iostream>using namespace std;//标号0为超级源点，标号1-26为26中应用程序，27-46(因为每个computer每天只能做一个程序，所以为computer是有容量的，要拆成两个点）computer的编号，48为超级汇点const int INF = 1000000;const int N = 100;int cap[N][N], flow[N][N], a[N], p[N];int s, t, sum;char ap, num, com[N], ans[15];bool maxFlow() {    queue <int> q;    memset(flow, 0, sizeof(flow));    int f = 0;    while ( 1 ) {        for ( int i = 0; i <= t; p[i++] = -1 );        memset(a, 0, sizeof(a));        a[s] = INF;        q.push(s);        while ( !q.empty() ) {            int u = q.front(); q.pop();            for ( int v = 0; v <= t; ++v ) {                if ( !a[v] && cap[u][v] > flow[u][v] ) {                    p[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 = p[u] ) {            if ( u < 37 && u > 26 ) if ( p[u] != -1 ) ans[u-27] = p[u] + 'A' - 1;            flow[p[u]][u] += a[t];            flow[u][p[u]] -= a[t];        }        f += a[t];    }    if ( f >= sum ) return true;    else return false;}void build () {    int u = ap-'A'+1, c = num-'0';    cap[s][u] += c;    for ( int v = 0; com[v] != ';'&& v < strlen(com); ++v ){         cap[u][com[v]-'0'+27] += 1;     }}void init() {    for ( int i = 27; i < 37; ++i ) {        cap[i][i+10] += 1; cap[i+10][t] += INF;    }} int main(){// freopen("input.txt","r", stdin);// freopen("output.txt", "w", stdout);     while ( cin >> ap >> num >> com ) {        memset(cap, 0, sizeof(cap));        getchar();        build();        s = 0, t = 47, sum = num-'0';        init();        while ( ( ap = getchar() ) != 0xa && ap != EOF ) {            scanf("%c %s", &num, com);            getchar();            sum += num - '0';            build();        }        for ( int i = 0; i < 10; ans[i++] = '_' );        if ( !maxFlow() ) cout << "!" << endl;        else {            for ( int i = 0; i < 10; ++i )                cout << ans[i];            cout << endl;        }        //for ( int i = 0; i <= t; printf("\n"), ++i )           //  for ( int j = 0; j <= t; ++j ) printf("%d ", cap[i][j] );    }}