HDU 3416

Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2470 Accepted Submission(s): 742



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

starvae@HDU

Source

HDOJ Monthly Contest – 2010.06.05

题意：
给出一张无向图，要求图上的不相交得最短路条数。

思路：
先求出单源最短路，从终点往起点倒着搜出最短路上的边（可以用双向链表），然后另这些边的容量为1建新图，跑一遍最大流即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

#define N 1010
#define M 101000
const int INF = 1000000000;

int n, m, sp, tp, cnte, cnte2, head
, lowdis
, head2
, dep
, tail
;
bool vis
;

class Edge
{
public:
int u, v, w, next, next2;
};
Edge edge1[M*2], edge2[M*2];

void addEdge1(int u, int v, int w)
{
edge1[cnte].u = u; edge1[cnte].v = v;
edge1[cnte].w = w; edge1[cnte].next = head[u];
head[u] = cnte;
edge1[cnte].next2 = tail[v];
tail[v] = cnte++;
}
void addEdge2(int u, int v, int w)
{
edge2[cnte2].u = u; edge2[cnte2].v = v;
edge2[cnte2].w = w; edge2[cnte2].next = head2[u];
head2[u] = cnte2++;

edge2[cnte2].v = u; edge2[cnte2].u = v;
edge2[cnte2].w = 0; edge2[cnte2].next = head2[v];
head2[v] = cnte2++;
}

int spfa()
{
queue<int> q;
for(int i = 1; i <= n; i++)
{
vis[i] = 0;
lowdis[i] = INF;
}
vis[sp] = 1; lowdis[sp] = 0;
q.push(sp);
while(!q.empty())
{
int cur = q.front();
q.pop(); vis[cur] = 0;
for(int i = head[cur]; i != -1; i = edge1[i].next)
{
int v = edge1[i].v;
if(lowdis[v] > lowdis[cur]+edge1[i].w)
{
lowdis[v] = lowdis[cur]+edge1[i].w;

if(!vis[v])
{
vis[v] = 1;
q.push(v);
}
}
}
}
return lowdis[tp] != INF;
}

void dfs(int cur)
{
for(int i = tail[cur]; i != -1; i = edge1[i].next2)
{
int u = edge1[i].u;
if(lowdis[u]+edge1[i].w == lowdis[cur])
{
addEdge2(u, cur, 1);
if(!vis[u])
{
vis[u] = 1;
dfs(u);
}
}
}
}

bool bfs()
{
memset(vis, 0, sizeof vis);
memset(dep, -1, sizeof dep);
queue<int> q;
q.push(sp);
vis[sp] = 1;
dep[sp] = 1;
while(!q.empty())
{
int cur = q.front();
q.pop();

for(int i = head2[cur]; i != -1; i = edge2[i].next)
{
int v = edge2[i].v;
if(!vis[v] && edge2[i].w > 0)
{
dep[v] = dep[cur]+1;
vis[v] = 1;
q.push(v);
}
}
}

return dep[tp] != -1;
}

int dfs2(int cur, int flow)
{
if(cur == tp) return flow;
int res = 0;
for(int i = head2[cur]; i != -1 && flow > res; i = edge2[i].next)
{
int v = edge2[i].v;
if(edge2[i].w > 0 && dep[v] == dep[cur]+1)
{
int x = min(edge2[i].w, flow-res);
int f = dfs2(v, x);
edge2[i].w-=f;
edge2[i^1].w+=f;
res += f;
}
}

if(!res) dep[cur] = -1;
return res;
}

int dinic()
{
int res = 0;
while(bfs())
{
int t;
while(t = dfs2(sp, INF))
{
res += t;
}
}
return res;
}

int main()
{
//freopen("C:\\Users\\Admin\\Desktop\\in.txt", "r", stdin);
int t; scanf("%d", &t);
while(t--)
{
cnte = 0;
cnte2 = 0;
memset(head, -1, sizeof head);
memset(head2, -1, sizeof head2);
memset(tail, -1, sizeof tail);
int u, v, w;
scanf("%d%d", &n, &m);
for(int i = 0; i < m; i++)
{
scanf("%d%d%d", &u, &v, &w);
addEdge1(u, v, w);
}
scanf("%d%d", &sp, &tp);
int ans;
if(!spfa()) ans = 0;
else
{
dfs(tp);
ans = dinic();
}
printf("%d\n", ans);
}
return 0;
}