http://poj.org/problem?id=1459
也是网络流的基础,只是虚拟出一个源点和终点,对应的生产值和消费值就加到与源点和终点的边上,然后做一次bfs就好了。
#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#define INF 999999999
//#define OPEN_FILE
using namespace std; const int MAXN = ;
int cap[MAXN][MAXN], flow[MAXN][MAXN], p[MAXN], c[MAXN], f[MAXN];
int n, ans; void get_cap(char s[])
{
int i, u, v, w;
u = ;
for (i = ; s[i] != ','; i++){
u = u * + (s[i] - '');
}
v = ;
for (++i; s[i] != ')'; i++){
v = v * + (s[i] - '');
}
int len = strlen(s);
w = ;
for (++i; i < len; i++){
w = w * + (s[i] - '');
}
cap[u][v] = w;
} void get_pc(char s[], int* t, int o)
{
int i, u, w;
u = ;
for (i = ; s[i] != ')'; i++){
u = u * + (s[i] - '');
}
int len = strlen(s);
w = ;
for (++i; i < len; i++){
w = w * + (s[i] - '');
}
if (o){
cap[n][u] = w;
}
else{
cap[u][n + ] = w;
} } void ek_bfs(){
queue<int> d;
int s, t, u, v;
int a[MAXN];
s = n;
t = n + ;
ans = ;
while (){
memset(a, , sizeof(a));
a[s] = INF;
d.push(s);
while (!d.empty()){
u = d.front();
d.pop();
for (v = ; v <= n + ; v++){
if (a[v] || cap[u][v] <= flow[u][v]) continue;
f[v] = u;
d.push(v);
a[v] = min(a[u] , cap[u][v] - flow[u][v]);
}
}
if (a[t] == ) break;
for (u = t; u != s; u = f[u]){
flow[f[u]][u] += a[t];
flow[u][f[u]] -= a[t];
}
ans += a[t];
}
} int main()
{
#ifdef OPEN_FILE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif // OPEN_FILE
int m, np, nc, i, j;
char s[];
while (~scanf("%d%d%d%d", &n, &np, &nc, &m)){
memset(cap, , sizeof(cap));
memset(flow, , sizeof(flow));
memset(p, , sizeof(p));
memset(c, , sizeof(c));
for (i = ; i <= m; i++){
scanf("%s", s);
get_cap(s);
}
for (i = ; i <= np; i++){
scanf("%s", s);
get_pc(s, p, );
}
for (i = ; i <= nc; i++){
scanf("%s", s);
get_pc(s, c, );
}
ek_bfs();
printf("%d\n", ans);
}
}