【ZJOI2010】【最大流】【最小费用最大流】网络扩容

第一问就是一个最大流的裸题。

第二问的话思维难度也不大，一开始加边的时候费用为0，然后如果原图有边，就加入一条对应的流量无限，费用为扩容费用的边，要注意的是如果有重边需要选择费用最小的一条，再加一个源点s，连边s-t，流量为k，费用为0的边，然后做一次最小费用最大流就可以了。

代码:

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1000 + 10;
const int maxque = 300000;
const int maxm = 200000;
struct pnode
{
int d,c,w,pos;
int next;
}E[maxm];
int cur[maxn],dis[maxn],gap[maxn];
int head[maxn],pre[maxn],que[maxque];
int map[maxn][maxn];
bool vis[maxn],flag[maxn][maxn];
int NE = 0;
int n,m,k;
int s,t,nodenum;
void init()
{
freopen("bzoj1834.in","r",stdin);
freopen("bzoj1834.out","w",stdout);
}

void insert(int u,int v,int c,int w)
{
E[NE].c = c;E[NE].w = w;E[NE].pos = v;E[NE].d = u;
E[NE].next = head[u];head[u] = NE++;
E[NE].c = 0;E[NE].w = -w;E[NE].pos = u;E[NE].d = v;
E[NE].next = head[v];head[v] = NE++;
}

void readdata()
{
memset(map,0x3f,sizeof(map));
memset(flag,false,sizeof(flag));
memset(E,0,sizeof(E));
memset(head,-1,sizeof(head));
s = 1,t = n,nodenum = n;
scanf("%d%d%d",&n,&m,&k);
for(int i = 1;i <= m;i++)
{
int u,v,c,w;
scanf("%d%d%d%d",&u,&v,&c,&w);
flag[u][v] = true;
map[u][v] = min(map[u][v],w);
insert(u,v,c,0);
}
}

inline void checkmin(int &a,int b)
{
if(a == -1 || a > b)a = b;
}

int sap()
{
memset(dis,0,sizeof(dis));
memset(gap,0,sizeof(gap));
for(int i = 1;i <= nodenum;i++)cur[i] = head[i];
int u = pre[s] = s,maxflow = 0,aug = -1;
gap[0] = nodenum;
while(dis[s] < nodenum)
{
loop:   for(int &i = cur[u];i != -1;i = E[i].next)
{
int v = E[i].pos;
if(E[i].c && dis[u] == dis[v] + 1)
{
checkmin(aug,E[i].c);
pre[v] = u;
u = v;
if(v == t)
{
maxflow += aug;
for(u = pre[u];v != s;v = u,u = pre[u])
{
E[cur[u]].c -= aug;
E[cur[u]^1].c += aug;
}
aug = -1;
}
goto loop;
}
}
int mind = nodenum;
for(int i = head[u];i != -1;i = E[i].next)
{
int v = E[i].pos;
if(E[i].c && (mind > dis[v]))
{
cur[u] = i;
mind = dis[v];
}
}
if((--gap[dis[u]]) == 0)break;
gap[dis[u] = mind + 1]++;
u = pre[u];
}
return maxflow;
}

int spfa()
{
int l,r;
memset(pre,-1,sizeof(pre));
memset(vis,0,sizeof(vis));
memset(dis,0x3f,sizeof(dis));
dis[s] = 0;
l = 0,r = 0;
que[r++] = s;
vis[s] = true;
while(l < r)
{
int u = que[l++];l %= maxque;
vis[u] = false;
for(int i = head[u];i != -1;i = E[i].next)
{
int v = E[i].pos;
if(E[i].c && dis[u] + E[i].w < dis[v])
{
dis[v] = dis[u] + E[i].w;
pre[v] = i;
if(!vis[v])
{
vis[v] = true;
que[r++] = v;r %= maxque;
}
}
}
}
if(dis[t] == inf)return false;
else return true;
}

int mcmf()
{
int ret = 0,flow = 0;
while(spfa())
{
int u = t;
int min = inf;
while(u != s)
{
if(E[pre[u]].c < min)min = E[pre[u]].c;
u = E[pre[u]].d;
}
flow += min;
u = t;
while(u != s)
{
E[pre[u]].c -= min;
E[pre[u]^1].c += min;
u = E[pre[u]].d;
}
ret += dis[t] * min;
}
return ret;
}

void solve()
{
s = 1,t = n,nodenum = n;
printf("%d ",sap());
insert(0,1,k,0);
for(int u = 1;u <= n;u++)
{
for(int v = 1;v <= n;v++)
{
if(flag[u][v])insert(u,v,inf,map[u][v]);
}
}
s = 0,t = n;
printf("%d\n",mcmf());
}

int main()
{
init();
readdata();
solve();
return 0;
}