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
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int INF = 0x3f3f3f3f;
4 const int maxn = 1010;
5 struct arc{
6 int to,w,next;
7 arc(int x = 0,int y = 0,int z = -1){
8 to = x;
9 w = y;
10 next = z;
11 }
12 }e[1000005];
13 int head[maxn],hd[maxn],d[maxn],gap[maxn],tot,n,m,S,T;
14 bool in[maxn] = {};
15 void add(int head[maxn],int u,int v,int wa,int wb){
16 e[tot] = arc(v,wa,head[u]);
17 head[u] = tot++;
18 e[tot] = arc(u,wb,head[v]);
19 head[v] = tot++;
20 }
21 void spfa(){
22 queue<int>q;
23 memset(d,0x3f,sizeof d);
24 d[S] = 0;
25 q.push(S);
26 while(!q.empty()){
27 int u = q.front();
28 q.pop();
29 in[u] = false;
30 for(int i = hd[u]; ~i; i = e[i].next){
31 if(d[e[i].to] > d[u] + e[i].w){
32 d[e[i].to] = d[u] + e[i].w;
33 if(!in[e[i].to]){
34 in[e[i].to] = true;
35 q.push(e[i].to);
36 }
37 }
38 }
39 }
40 for(int i = 1; i <= n; ++i){
41 for(int j = hd[i]; ~j; j = e[j].next){
42 if(d[e[j].to] == d[i] + e[j].w)
43 add(head,i,e[j].to,1,0);
44 }
45 }
46 }
47 int dfs(int u,int low){
48 if(u == T) return low;
49 int tmp = 0,minH = n - 1;
50 for(int i = head[u]; ~i; i = e[i].next){
51 if(e[i].w){
52 if(d[e[i].to] + 1 == d[u]){
53 int a = dfs(e[i].to,min(low,e[i].w));
54 e[i].w -= a;
55 e[i^1].w += a;
56 tmp += a;
57 low -= a;
58 if(!low) break;
59 if(d[S] >= n) return tmp;
60 }
61 if(e[i].w) minH = min(minH,d[e[i].to]);
62 }
63 }
64 if(!tmp){
65 if(--gap[d[u]] == 0) d[S] = n;
66 ++gap[d[u] = minH + 1];
67 }
68 return tmp;
69 }
70 void bfs(){
71 queue<int>q;
72 memset(d,-1,sizeof d);
73 memset(gap,0,sizeof gap);
74 q.push(T);
75 d[T] = 0;
76 while(!q.empty()){
77 int u = q.front();
78 q.pop();
79 ++gap[d[u]];
80 for(int i = head[u]; ~i; i = e[i].next){
81 if(d[e[i].to] == -1){
82 d[e[i].to] = d[u] + 1;
83 q.push(e[i].to);
84 }
85 }
86 }
87 }
88 int sap(int ret = 0){
89 bfs();
90 while(d[S] < n) ret += dfs(S,INF);
91 return ret;
92 }
93 int main(){
94 int kase,u,v,w;
95 scanf("%d",&kase);
96 while(kase--){
97 scanf("%d%d",&n,&m);
98 memset(head,-1,sizeof head);
99 memset(hd,-1,sizeof hd);
100 for(int i = tot = 0; i < m; ++i){
101 scanf("%d%d%d",&u,&v,&w);
102 if(u == v) continue;
103 add(hd,u,v,w,INF);
104 }
105 scanf("%d%d",&S,&T);
106 spfa();
107 printf("%d\n",sap());
108 }
109 return 0;
110 }