题目链接:click here
题意:一个电力网络有n个点,有np个发电站,nc个消耗点,其余的为中转站。m条电缆,中转站既不发电也不耗电。每条电缆都有一个最大容量。
思路:设置一个超级源点和一个超级汇点,将所有的源点和汇点分别放进去,Dinic 算法实现。
注意括号的处理。
代码:
#include <math.h>
#include <queue>
#include <deque>
#include <vector>
#include <stack>
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define mem(a,b) memset(a,b,sizeof(a))
int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
const double eps = 1e-6;
const double Pi = acos(-1.0);
const int maxn = 505;
const int inf=0x3f3f3f3f;
struct edge
{
int to, cap, rev;
};
vector <edge> G[maxn];
int level[maxn]; //顶点到源点的距离编号
int iter[maxn]; //当前弧,在其之前的边不用再考虑
int n,np,nc,m,i,j,u,v,z;
void add_edge(int from, int to, int cap)
{
G[from].push_back((edge)
{
to, cap, G[to].size()
});
G[to].push_back((edge)
{
from, 0, G[from].size()-1
});
}
//bfs用来计算从源点出发所有点的距离编号
void bfs(int s)
{
memset(level, -1, sizeof(level));
queue<int> que;
level[s] = 0;
que.push(s);
while(!que.empty())
{
int v = que.front();
que.pop();
for (int i = 0; i < G[v].size(); i++)
{
edge &e = G[v][i];
if (e.cap > 0 && level[e.to] < 0)
{
level[e.to] = level[v] + 1;
que.push(e.to);
}
}
}
}
//通过DFS寻找当前的最短的增广路
int dfs(int v, int t, int f)
{
if (v == t) return f;
for (int &i = iter[v]; i < G[v].size(); i++)
{
edge &e = G[v][i];
if (e.cap > 0 && level[v] < level[e.to])
{
int d = dfs(e.to, t, min(f, e.cap));
if (d > 0)
{
e.cap-=d;
G[e.to][e.rev].cap += d;
return d;
}
}
}
return 0;
}
void input()
{
char str[100];
int a, b, c;
int s = n;
int t = n + 1;
for (int i = 0; i < m; i++)
{
scanf("%[^(]", str);
scanf("(%d,%d)%d", &a, &b, &c);
add_edge(a, b, c);
}
for (int i = 0; i <np; i++)
{
scanf("%[^(]", str);
scanf("(%d)%d", &a, &c);
add_edge(s, a, c);
}
for (int i = 0; i < nc; i++)
{
scanf("%[^(]", str);
scanf("(%d)%d", &a, &c);
add_edge(a, t, c);
}
}
int max_flow(int s, int t)
{
int flow = 0;
for(;;)
{
bfs(s);
if (level[t] < 0) return flow;
memset(iter, 0, sizeof(iter));
int f;
while (( f = dfs(s, t, inf)) > 0)
{
flow += f;
}
}
}
int main()
{
while (~scanf("%d%d%d%d", &n, &np, &nc, &m))
{
char str;
for (i = 0; i < maxn; i++) G[i].clear();
input();
printf("%d\n", max_flow(n, n+1));
}
return 0;
}