Codeforces 362E Petya and Pipes 费用流建图

题意：

给一个网络中某些边增加容量，增加的总和最大为K，使得最大流最大。

费用流:在某条边增加单位流量的费用。

那么就可以2个点之间建2条边，第一条给定边(u,v,x,0)这条边费用为0

同时另一条边(u,v,K,1)费用为1，那么就可以通过限制在增广时相应的费用即可找出最大流

个人觉得这样做的原因是每次增光都是最优的。所以通过限制最终费用不超过K可以得到最优解

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
const int MAXN = 110;
const int INF = 0x3f3f3f3f;
struct node
{
int u,v,next;
int flow,cap,cost;
}edge[MAXN * MAXN * 4];
int cnt,src,tag;
int C,F;
int K,N;
queue<int>q;
bool inq[MAXN];int d[MAXN];
int head[MAXN],p[MAXN];
int tot = 0;

void init()
{
memset(head,-1,sizeof(head));
tot = 0;
}

void add_edge(int u,int v,int cap,int cost)
{
edge[cnt].u = u;
edge[cnt].v = v;
edge[cnt].cap = cap;
edge[cnt].flow = 0;
edge[cnt].cost = cost;
edge[cnt].next = head[u];
head[u] = cnt++;
//反向
edge[cnt].v = u;
edge[cnt].u = v;
edge[cnt].flow = 0;
edge[cnt].cap = 0;
edge[cnt].cost = - cost;
edge[cnt].next = head[v];
head[v] = cnt++;
}

bool SPFA(int s, int t)
{
while (!q.empty()) q.pop();
memset(inq,false,sizeof(inq));
memset(d,0x3f,sizeof(d));
memset(p,-1,sizeof(p));
d[s] = 0;
q.push(s);
inq[s] = true;
while (!q.empty())
{
int u = q.front(); q.pop();
inq[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v;
if (d[v] > d[u] + edge[i].cost && edge[i].cap > edge[i].flow)
{
d[v] = d[u] + edge[i].cost;
p[v] = i;
if (!inq[v])
{
q.push(v);
inq[v] = true;
}
}
}
}
if(d[tag] == INF) return false;
int a = INF;
for (int i = p[tag]; i != -1; i = p[edge[i].u])
a = min(a,edge[i].cap - edge[i].flow);
if(C + d[tag] * a > K)
{
F += (K - C) / d[tag];
return false;
}
return true;
}
void slove()
{
C = F = 0;
while(SPFA(src,tag))
{
int a = INF;
for (int i = p[tag]; i != -1; i = p[edge[i].u])
a = min(a,edge[i].cap - edge[i].flow);
for (int i = p[tag]; i != -1; i = p[edge[i].u])
{
edge[i].flow += a;
edge[i ^ 1].flow -= a;
}
C += d[tag] * a;
F += a;
}
}

int main()
{
while (scanf("%d%d",&N,&K) != EOF)
{
init();
for (int i = 1 ; i <= N ; i++)
for (int j = 1 ; j <= N ; j++)
{
int x;
scanf("%d",&x);
if (x)
{
add_edge(i,j,x,0);
add_edge(i,j,K,1);
}
}
src = 1;
tag = N;
slove();
printf("%d\n",F);
}
return 0;
}