最小费用最大流之 zkw费用流与普通费用流
这个是原作者的博文地址
这是今天刚学的,不过理解上还是很浅薄。最近发现算法不能融会贯通还是因为自己太死板了。
奉上一个基础版本的模板, POJ 2195 的代码。
本模板是不能直接用于任何有负权的图,更不能用于有负圈的情况
[cpp]
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 222
#define MAXM 55555
#define INF 1000000007
using namespace std;
struct EDGE
{
int cost, cap, v;
int next, re;
}edge[MAXM];
int head[MAXN], e;
int vis[MAXN];
int ans, cost, src, des, n;
void init()
{
memset(head, -1, sizeof(head));
e = 0;
ans = cost = 0;
}
void add(int u, int v, int cap, int cost)
{
edge[e].v = v;
edge[e].cap = cap;
edge[e].cost = cost;
edge[e].re = e + 1;
edge[e].next = head[u];
head[u] = e++;
edge[e].v = u;
edge[e].cap = 0;
edge[e].cost = -cost;
edge[e].re = e - 1;
edge[e].next = head[v];
head[v] = e++;
}
int aug(int u, int f)
{
if(u == des)
{
ans += cost * f;
return f;
}
vis[u] = 1;
int tmp = f;
for(int i = head[u]; i != -1; i = edge[i].next)
if(edge[i].cap && !edge[i].cost && !vis[edge[i].v])
{
int delta = aug(edge[i].v, tmp < edge[i].cap ? tmp : edge[i].cap);
edge[i].cap -= delta;
edge[edge[i].re].cap += delta;
tmp -= delta;
if(!tmp) return f;
}
return f - tmp;
}
bool modlabel()
{
int delta = INF;
for(int u = 1; u <= n; u++)
if(vis[u])
for(int i = head[u]; i != -1; i = edge[i].next)
if(edge[i].cap && !vis[edge[i].v] && edge[i].cost < delta) delta = edge[i].cost;
if(delta == INF) return false;
for(int u = 1; u <= n; u++)
if(vis[u])
for(int i = head[u]; i != -1; i = edge[i].next)
edge[i].cost -= delta, edge[edge[i].re].cost += delta;
cost += delta;
return true;
}
void costflow()
{
do
{
do
{
memset(vis, 0, sizeof(vis));
}while(aug(src, INF));
}while(modlabel());
}
int nt, m;
struct point
{
int x, y;
}p[MAXN], h[MAXN];
int d[MAXN][MAXN];
char s[MAXN][MAXN];
int main()
{
while(scanf("%d%d", &m, &nt) != EOF)
{
if(m == 0 && nt == 0) break;
for(int i = 0; i < m; i++)
scanf("%s", s[i]);
int hcnt = 0, pcnt = 0;
for(int i = 0; i < m; i++)
for(int j = 0; j < nt; j++)
{
if(s[i][j] == 'H')
{
hcnt++;
h[hcnt].x = i;
h[hcnt].y = j;
}
else if(s[i][j] == 'm')
{
pcnt++;
p[pcnt].x = i;
p[pcnt].y = j;
}
}
for(int i = 1; i <= pcnt; i++)
for(int j = 1; j <= hcnt; j++)
d[i][j] = abs(p[i].x - h[j].x) + abs(p[i].y - h[j].y);
init();
n = hcnt + pcnt + 2;
补充:软件开发 , C++ ,