Marriage Match IV
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2275 Accepted Submission(s): 691
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<c<=1000)it means there is a road from a to b and it‘s distance is c, while there may have no road from b to a. There may have a road from a to a,but you can ignore it. If there are two roads
from a to b, they are different.
At last is a line with two integer A and B(1<=A,B<=N,A!=B), means the number of city A and city B.
There may be some blank line between each case.
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
Author
[email protected]
Source
HDOJ Monthly Contest – 2010.06.05
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题意：有n个城市m条边，告诉起点和终点，问从起点到终点不走重复边的最短路有多少条。
思路：求最短路的条数可以用最大流，建图时要去掉不是最短路上的边，判断某条边是不是最短路上的边：如果满足dist1[u] + dist2[v] + E1[i].len = dist1[End]则可以说明该边是最短路上的边，其中 dist1[] 为各点到起点的最短距离，dist2[] 为各点到终点的最短距离。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define MAXN 1005
#define MAXM 300005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
struct Edge
{
int to,next,cap,flow;
}edge[MAXM];
int n,m,start,End;
int tol,tol1,tol2;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
struct EDGE
{
int u,v,len,next;
}E1[MAXM],E2[MAXM];
int head1[MAXN],dist1[MAXN],inq[MAXN];
int head2[MAXN],dist2[MAXN];
void init()
{
tol=0;tol1=0;tol2=0;
memset(head,-1,sizeof(head));
memset(head1,-1,sizeof(head1));
memset(head2,-1,sizeof(head2));
}
void add1(int u,int v,int w)
{
E1[tol1].u=u; E1[tol1].v=v; E1[tol1].len=w; E1[tol1].next=head1[u]; head1[u]=tol1++;
}
void add2(int u,int v,int w)
{
E2[tol2].u=u; E2[tol2].v=v; E2[tol2].len=w; E2[tol2].next=head2[u]; head2[u]=tol2++;
}
//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++;
}
void spfa(int s,int *dist,int *head,EDGE *E) //spfa求最短路
{
memset(inq,0,sizeof(inq));
queue<int>Q;
Q.push(s);
inq[s]=1;
dist[s]=0;
while (!Q.empty())
{
int u=Q.front(); Q.pop();
inq[u]=0;
for (int i=head[u];i+1;i=E[i].next)
{
int v=E[i].v;
if (dist[v]>dist[u]+E[i].len)
{
dist[v]=dist[u]+E[i].len;
if (!inq[v])
{
inq[v]=1;
Q.push(v);
}
}
}
}
}
//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;
while (dep[start]<N)
{
if (u==End)
{
int Min=INF;
for (int 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 (int 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 (int 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 (int 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 i,j,t,u,v,w;
scanf("%d",&t);
while (t--)
{
init();
scanf("%d%d",&n,&m);
for (i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
if (u==v) continue; //去自环
add1(u,v,w);
add2(v,u,w);
}
scanf("%d%d",&start,&End);
memset(dist1,INF,sizeof(dist1));
memset(dist2,INF,sizeof(dist2));
spfa(start,dist1,head1,E1);
spfa(End,dist2,head2,E2);
for (i=0;i<tol1;i++) //建图
{
if (dist1[E1[i].u]+dist2[E1[i].v]+E1[i].len==dist1[End])
addedge(E1[i].u,E1[i].v,1);
}
printf("%d\n",sap(start,End,n));
}
return 0;
}