題意:看題看的心累,其實就是所有power station是源點,所有consumer是匯點,然後就是一個最大流問題了~
讀入寫的有點蠢,其實scanf(" ()")
這種就夠啦~
另外,對於單路增廣的dinic來說,當前弧優化真的很有必要。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
using namespace std;
const int maxn = 200 + 10, maxe = maxn*maxn;
const int inf = 0x3f3f3f3f;
struct node
{
int to,next,cap,rev;
node(){}
node(int a,int b,int c,int d){to = a; next = b; cap = c; rev = d;}
}edge[maxe << 2];
int num;
int h[maxn],deg[maxn],cur[maxn];
int n,np,nc,m;
int s,t; //源, 匯
void init()
{
for(int i = 0; i < maxn; i++)
h[i] = -1,deg[i] = 0;
s = 0, t = n+1;
num = 0;
}
void add(int u,int v,int cap)
{
edge[num] = node(v,h[u],cap,num+1); h[u] = num++;
edge[num] = node(u,h[v],0,num-1); h[v] = num++;
}
int bfs()
{
for(int i = 0; i < maxn; i++) deg[i] = -1;
queue<int> q;
q.push(0);
deg[0] = 0;
while(!q.empty())
{
int u = q.front(); q.pop();
for(int i = h[u] ; ~i ; i = edge[i].next)
{
int v = edge[i].to,cap = edge[i].cap;
if(cap && deg[v] == -1)
{
deg[v] = deg[u] + 1;
q.push(v);
}
}
}
return deg[t] != -1;
}
int dfs(int u,int f)
{
if(u == t) return f;
for(int i = cur[u]; ~i; i = edge[i].next)
{
cur[u] = i;
int v = edge[i].to,cap = edge[i].cap, rev = edge[i].rev;
if(cap && deg[v] == deg[u] + 1)
{
int k = dfs(v,min(f,cap));
if(k == 0) continue;
edge[i].cap -= k;
edge[rev].cap += k;
return k;
}
}
return 0;
}
void solve()
{
int flow = 0;
while(bfs())
{
for(int i = 0; i < maxn; i++)
cur[i] = h[i];
for(;;)
{
int k = dfs(0,inf);
if(k == 0) break;
flow += k;
}
}
printf("%d\n",flow);
}
char str[1000];
void readnum(int way)
{
int v = 0,cap = 0;
scanf("%s",str); int slen = strlen(str);
int j;
for(j = 1; j < slen; j++)
{
if(str[j] == ')') break;
v = v*10 + str[j] - '0';
}
j++;
for(; j < slen; j++)
cap = cap*10 + str[j] - '0';
if(way) add(s,v+1,cap);
else add(v+1,t,cap);
}
void read()
{
for(int i = 0; i < m; i++)
{
int u = 0,v = 0,cap = 0;
scanf("%s",str);
int slen = strlen(str);
int j;
for(j = 1; j < slen;j++)
{
if(str[j] == ',') break;
u = u*10 + str[j] - '0';
}
j++;
for(; j < slen; j++)
{
if(str[j] == ')') break;
v = v*10 + str[j] - '0';
}
j++;
for(; j < slen; j++)
cap = cap*10 + str[j] - '0';
// cout << u << " " << v << " " <<cap << endl;
add(u+1,v+1,cap);
}
for(int i = 0; i < np; i++)
readnum(1);
for(int i = 0; i < nc; i++)
readnum(0);
}
int main()
{
//freopen("D:\\in.txt","r",stdin);
int a,b,c;
while(~scanf("%d",&n))
{
init();
scanf("%d%d%d",&np,&nc,&m);
read();
solve();
}
return 0;
}