HDU 3416 Marriage Match IV

Marriage Match IV

Time Limit: 1000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 3416
64-bit integer IO format: %I64d Java class name: Main
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

Source

HDOJ Monthly Contest – 2010.06.05

解题：直接把所有的最短路拿出来跑最小割

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1010;
struct arc{
int to,w,next;
arc(int x = 0,int y = 0,int z = -1){
to = x;
w = y;
next = z;
}
}e[1000005];
int head[maxn],hd[maxn],d[maxn],gap[maxn],tot,n,m,S,T;
bool in[maxn] = {};
void add(int head[maxn],int u,int v,int wa,int wb){
e[tot] = arc(v,wa,head[u]);
head[u] = tot++;
e[tot] = arc(u,wb,head[v]);
head[v] = tot++;
}
void spfa(){
queue<int>q;
memset(d,0x3f,sizeof d);
d[S] = 0;
q.push(S);
while(!q.empty()){
int u = q.front();
q.pop();
in[u] = false;
for(int i = hd[u]; ~i; i = e[i].next){
if(d[e[i].to] > d[u] + e[i].w){
d[e[i].to] = d[u] + e[i].w;
if(!in[e[i].to]){
in[e[i].to] = true;
q.push(e[i].to);
}
}
}
}
for(int i = 1; i <= n; ++i){
for(int j = hd[i]; ~j; j = e[j].next){
if(d[e[j].to] == d[i] + e[j].w)
add(head,i,e[j].to,1,0);
}
}
}
int dfs(int u,int low){
if(u == T) return low;
int tmp = 0,minH = n - 1;
for(int i = head[u]; ~i; i = e[i].next){
if(e[i].w){
if(d[e[i].to] + 1 == d[u]){
int a = dfs(e[i].to,min(low,e[i].w));
e[i].w -= a;
e[i^1].w += a;
tmp += a;
low -= a;
if(!low) break;
if(d[S] >= n) return tmp;
}
if(e[i].w) minH = min(minH,d[e[i].to]);
}
}
if(!tmp){
if(--gap[d[u]] == 0) d[S] = n;
++gap[d[u] = minH + 1];
}
return tmp;
}
void bfs(){
queue<int>q;
memset(d,-1,sizeof d);
memset(gap,0,sizeof gap);
q.push(T);
d[T] = 0;
while(!q.empty()){
int u = q.front();
q.pop();
++gap[d[u]];
for(int i = head[u]; ~i; i = e[i].next){
if(d[e[i].to] == -1){
d[e[i].to] = d[u] + 1;
q.push(e[i].to);
}
}
}
}
int sap(int ret = 0){
bfs();
while(d[S] < n) ret += dfs(S,INF);
return ret;
}
int main(){
int kase,u,v,w;
scanf("%d",&kase);
while(kase--){
scanf("%d%d",&n,&m);
memset(head,-1,sizeof head);
memset(hd,-1,sizeof hd);
for(int i = tot = 0; i < m; ++i){
scanf("%d%d%d",&u,&v,&w);
if(u == v) continue;
add(hd,u,v,w,INF);
}
scanf("%d%d",&S,&T);
spfa();
printf("%d\n",sap());
}
return 0;
}

View Code