题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549
2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1
Case 1: 1 Case 2: 2
代码如下:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 32;//点数的最大值
const int MAXM = 1017;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
int to, cap, flow;
int next;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int dep[MAXN],pre[MAXN],cur[MAXN];
int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为y
void init()
{
tol = 0;
memset(head,-1,sizeof (head));
}
//加边,单向图三个参数,双向图四个参数
void addedge (int u,int v,int w,int rw=0)
{
edge[tol].to = v;edge[tol].cap = w;
edge[tol].next = head[u];
edge[tol].flow = 0;
head[u] = tol++;
edge[tol].to = u;edge[tol].cap = rw;
edge[tol]. next = head[v];
edge[tol].flow = 0;head[v]=tol++;
}
//输入参数:起点、终点、点的总数
//点的编号没有影响,只要输入点的总数
int sap(int start,int end, int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
int ans = 0;
int i;
while(dep[start] < N)
{
if(u == end)
{
int Min = INF;
for( i = pre[u];i != -1; i = pre[edge[i^1]. to])
{
if(Min > edge[i].cap - edge[i]. flow)
Min = edge[i].cap - edge[i].flow;
}
for( i = pre[u];i != -1; i = pre[edge[i^1]. to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for( i = cur[u]; i != -1;i = edge[i].next)
{
v = edge[i]. to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
{
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if(flag)
{
u = v;
continue;
}
int Min = N;
for( i = head[u]; i != -1; i = edge[i]. next)
{
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
{
Min = dep[edge[i].to];
cur[u] = i;
}
}
gap[dep[u]]--;
if(!gap[dep[u]])
return ans;
dep[u] = Min+1;
gap[dep[u]]++;
if(u != start)
u = edge[pre[u]^1].to;
}
return ans;
}
int main()
{
int n, m;
int a, b, w;
int c, s, t;
int i;
int T;
int cas = 0;
scanf("%d",&T);
while(T--)
{
init();//初始化
scanf("%d%d",&n,&m);
for(i = 1; i <= m; i++)//边数
{
scanf("%d%d%d",&a,&b,&w);
addedge(a,b,w,0);
// addedge(b,a,w,0);
}
int ans = sap(1, n, n);
printf("Case %d: %d\n",++cas,ans);
}
return 0;
}hdu 3549 Flow Problem(最大流模板题),布布扣,bubuko.com
原文地址:http://blog.csdn.net/u012860063/article/details/38467299