ZOJ 1760 How Many Shortest Path（最短路+网络流之最大流）

题目地址：ZOJ 1760

利用这篇博客里提到的判断最短路边的方法，标记最短路的边，然后跑一次最大流就行了。不多说了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include<algorithm>

using namespace std;
const int INF=0x3f3f3f3f;
int head[200], source, sink, nv, cnt;
int cur[200], num[200], d[200], pre[200];
struct node
{
int u, v, cap, next;
} edge[1000000];
void add(int u, int v, int cap)
{
edge[cnt].v=v;
edge[cnt].cap=cap;
edge[cnt].next=head[u];
head[u]=cnt++;

edge[cnt].v=u;
edge[cnt].cap=0;
edge[cnt].next=head[v];
head[v]=cnt++;
}
void bfs()
{
memset(d,-1,sizeof(d));
memset(num,0,sizeof(num));
queue<int>q;
q.push(sink);
d[sink]=0;
num[0]=1;
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(d[v]==-1)
{
d[v]=d[u]+1;
num[d[v]]++;
q.push(v);
}
}
}
}
void isap()
{
memcpy(cur,head,sizeof(cur));
bfs();
int flow=0, u=pre[source]=source, i;
while(d[source]<nv)
{
if(u==sink)
{
int f=INF, pos;
for(i=source;i!=sink;i=edge[cur[i]].v)
{
if(f>edge[cur[i]].cap)
{
f=edge[cur[i]].cap;
pos=i;
}
}
for(i=source;i!=sink;i=edge[cur[i]].v)
{
edge[cur[i]].cap-=f;
edge[cur[i]^1].cap+=f;
}
flow+=f;
u=pos;
}
for(i=cur[u];i!=-1;i=edge[i].next)
{
if(d[edge[i].v]+1==d[u]&&edge[i].cap)
{
break;
}
}
if(i!=-1)
{
cur[u]=i;
pre[edge[i].v]=u;
u=edge[i].v;
}
else
{
if(--num[d[u]]==0) break;
int mind=nv;
for(i=head[u];i!=-1;i=edge[i].next)
{
if(mind>d[edge[i].v]&&edge[i].cap)
{
mind=d[edge[i].v];
cur[u]=i;
}
}
d[u]=mind+1;
num[d[u]]++;
u=pre[u];
}
}
printf("%d\n",flow);
}
int d1[200], vis[200], cnt1, head1[200];
struct node1
{
int u, v, w, next;
} path[1000000];
void add1(int u, int v, int w)
{
path[cnt1].v=v;
path[cnt1].w=w;
path[cnt1].next=head1[u];
head1[u]=cnt1++;
}
void spfa()
{
int i;
deque<int>q;
q.push_front(source);
memset(d1,INF,sizeof(d1));
memset(vis,0,sizeof(vis));
d1[source]=0;
while(!q.empty())
{
int u=q.front();
q.pop_front();
vis[u]=0;
for(i=head1[u]; i!=-1; i=path[i].next)
{
int v=path[i].v;
if(d1[v]>d1[u]+path[i].w)
{
d1[v]=d1[u]+path[i].w;
if(!vis[v])
{
vis[v]=1;
if(!q.empty()&&d1[v]<d1[q.front()])
{
q.push_front(v);
}
else
q.push_back(v);
}
}
}
}
}
int main()
{
int n, i, j, x;
while(scanf("%d",&n)!=EOF)
{
memset(head1,-1,sizeof(head1));
cnt1=0;
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
scanf("%d",&x);
if(x>=0&&i!=j)
{
add1(i,j,x);
}
}
}
scanf("%d%d",&source,&sink);
if(source==sink)
{
printf("inf\n");
continue ;
}
nv=n;
spfa();
memset(head,-1,sizeof(head));
cnt=0;
for(i=0;i<n;i++)
{
for(j=head1[i];j!=-1;j=path[j].next)
{
if(d1[path[j].v]==d1[i]+path[j].w)
{
add(i,path[j].v,1);
}
}
}
isap();
}
return 0;
}