HDU3416 Marriage Match IV(最短路，网络流，最大流，SPFA,ISAP算法)

Problem Description

Do not sincere non-interference。 Like that show, now starvae also take

part in a show, but it take place between city A and B. Starvae is in

city A and girls are in city B. Every time starvae can get to city B

and make a data with a girl he likes. But there are two problems with

it, one is starvae must get to B within least time, it’s said that he

must take a shortest path. Other is no road can be taken more than

once. While the city starvae passed away can been taken more than

once.

So, under a good RP, starvae may have many chances to get to city B.

But he don’t know how many chances at most he can make a data with the

girl he likes . Could you help starvae?

Input

The first line is an integer T indicating the case number.(1<=T<=65)

For each case,there are two integer n and m in the first line (

2<=n<=1000, 0<=m<=100000 ) ,n is the number of the city and m is the

number of the roads.

Then follows m line ,each line have three integers

a,b,c,(1<=a,b<=n,0

Output

Output a line with a integer, means the chances starvae can get at

most.

Sample Input

3
7 8
1 2 1
1 3 1
2 4 1
3 4 1
4 5 1
4 6 1
5 7 1
6 7 1
1 7

6 7
1 2 1
2 3 1
1 3 3
3 4 1
3 5 1
4 6 1
5 6 1
1 6

2 2
1 2 1
1 2 2
1 2

Sample Output

2
1
1

思路

首先
4000
是题意，一个人要从一个城市走到另一个城市，每次都走最短路，但是每次走的路不能是同一条，问一共有多少种不同的走法。

输入是这样给出的，首先是T组数据，然后是两个数n和m，代表有n个点和m条边。然后有m行数据，每行是三个数u,v,w,代表从u到v的权值是w。然后是两个数，代表了起点和终点.

首先我们的思路是，从图中剔除不是最短路的边，然后把剩下的边(最短路上的边)建图，然后边权为1，求最大流，求出的值就是有多少种走法。

那么问题来了，如何判断一条边(u,v)是不是最短路的边呢。

我们假定起点是

st

,终点为

ed

,当前要判断的边是(u,v),从起点开始的最短路数组为dis1[],从终点开始的最短路数组为dis2[],那么我们要判断边是不是在最短路上，只要：

dis1[u]+(u,v)+dis2[v]==dis1[ed]

成立就可以确定边在最短路上，我们根据这个条件，把在最短路上的边建图，求一下最大流就是答案

SAP算法的模板用了

kuangbin

的。

代码

#include <cstdio>
#include <cstring>
#include <cctype>
#include <stdlib.h>
#include <string>
#include <map>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
const int N=1000+20;
const int M=2*100000+20;
int top;
int h
,pre
,g
,first
,cur
;//h[i]记录每个节点的高度，pre[i]记录前驱,g[i]表示距离为i个节点数有多少个
struct node
{
int v,next,cap;
} E[M];
void init()
{
mem(first,-1);
top=0;
}
void add_edge(int u,int v,int c)
{
E[top].v=v;
E[top].cap=c;
E[top].next=first[u];
first[u]=top++;
E[top].v=u;
E[top].cap=0;
E[top].next=first[v];
first[v]=top++;
}
int sap(int start,int end,int nodenum)
{
memset(h,0,sizeof(h));
memset(g,0,sizeof(g));
memcpy(cur,first,sizeof(first));
int u=pre[start]=start,maxflow=0,aug=-1;
g[0]=nodenum;
while(h[start]<nodenum)
{
loop:
for(int &i=cur[u]; i!=-1; i=E[i].next)
{
int v=E[i].v;
if(E[i].cap&&h[u]==h[v]+1)
{
if(aug==-1||aug>E[i].cap)
aug=E[i].cap;
pre[v]=u;
u=v;
if(v==end)
{
maxflow+=aug;
for(u=pre[u]; v!=start; v=u,u=pre[u])
{
E[cur[u]].cap-=aug;
E[cur[u]^1].cap+=aug;
}
aug=-1;
}
goto loop;
}
}
int mindis=nodenum;
for(int i=first[u]; i!=-1; i=E[i].next)
{
int v=E[i].v;
if(E[i].cap&&mindis>h[v])
{
cur[u]=i;
mindis=h[v];
}
}
if((--g[h[u]])==0)break;
g[h[u]=mindis+1]++;
u=pre[u];
}
return maxflow;
}

int n,m;
struct SPFA
{
void init()
{
mem(first,-1);
mem(vis,0);
mem(dis,0);
len=1;
}
int first
,len,vis
,dis
;
struct node
{
int u,v,w,next;
} G[M];
void add_edge(int u,int v,int w)
{
G[len].v=v,G[len].w=w;
G[len].next=first[u];
first[u]=len++;
}
void spfa(int st)
{
for(int i=1; i<=n; i++)
{
dis[i]=inf;
vis[i]=0;
}
dis[st]=0;
vis[st]=1;
queue<int>q;
q.push(st);
while(!q.empty())
{
st=q.front();
q.pop();
vis[st]=0;
for(int i=first[st]; i!=-1; i=G[i].next)
{
int v=G[i].v,w=G[i].w;
if(dis[v]>dis[st]+w)
{
dis[v]=dis[st]+w;
if(!vis[v])
{
vis[v]=1;
q.push(v);
}
}
}
}
}
} ac1,ac2;
int main()
{
int t,st,ed,u,v,w;
scanf("%d",&t);
while(t--)
{
init();
ac1.init();
ac2.init();
scanf("%d%d",&n,&m);
for(int i=1; i<=m; i++)
{
scanf("%d%d%d",&u,&v,&w);
ac1.add_edge(u,v,w);
ac2.add_edge(v,u,w);
}
scanf("%d%d",&st,&ed);
ac1.spfa(st);
ac2.spfa(ed);
for(int i=1; i<=n; i++)
{
for(int j=ac1.first[i]; ~j; j=ac1.G[j].next)
{
u=i,v=ac1.G[j].v,w=ac1.G[j].w;
if(ac1.dis[u]+w+ac2.dis[v]==ac1.dis[ed])
add_edge(u,v,1);
}
}
printf("%d\n",sap(st,ed,n));
}
return 0;
}