Barricade
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2418 Accepted Submission(s): 696
Problem Description
The empire is under attack again. The general of empire is planning to defend his castle. The land can be seen as N towns and M roads, and each road has the same length and connects two towns. The town numbered 1 is where general's castle is located, and the town numbered N is where the enemies are staying. The general supposes that the enemies would choose a shortest path. He knows his army is not ready to fight and he needs more time. Consequently he decides to put some barricades on some roads to slow down his enemies. Now, he asks you to find a way to set these barricades to make sure the enemies would meet at least one of them. Moreover, the barricade on the i-th road requires wi units of wood. Because of lacking resources, you need to use as less wood as possible.
Input
The first line of input contains an integer t, then t test cases follow.
For each test case, in the first line there are two integers N(N≤1000) and M(M≤10000).
The i-the line of the next M lines describes the i-th edge with three integers u,v and w where 0≤w≤1000 denoting an edge between u and v of barricade cost w.
Output
For each test cases, output the minimum wood cost.
Sample Input
1 4 4 1 2 1 2 4 2 3 1 3 4 3 4
Sample Output
4
Source
2016 ACM/ICPC Asia Regional Qingdao Online
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#include <map>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1000 + 10;
const int maxm = 10000 + 10;
typedef pair<int,int> P;
int n,m;
int l[maxn];//记录层数
int h[maxn];//链式前向星
int cur[maxn];
int tot = 0;
struct edge
{
int to;
int c;
int next;
edge(int x = 0, int y = 0, int z = 0) : to(x), c(y), next(z) {}
}es[maxm*2];//记录边 注意是2倍
void add_edge(int u, int v, int c)
{
es[tot] = edge(v,c,h[u]);
h[u] = tot++;
}
bool bfs(int s, int t)
{
memset(l,0,sizeof(l));
l[s] = 1;
queue <int> q;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
if(u == t) return true;
for(int i = h[u]; i != -1; i = es[i].next)
{
int v = es[i].to;
if(!l[v] && es[i].c) {l[v] = l[u] + 1; q.push(v);}
}
}
return false;
}
int dfs(int x, int t, int mf)
{
if(x == t) return mf;
int ret = 0;
for(int &i = cur[x]; i != -1; i = es[i].next)
{
if(es[i].c && l[x] == l[es[i].to] - 1)
{
int f = dfs(es[i].to,t,min(es[i].c,mf - ret));
es[i].c -= f;
es[i^1].c += f;
ret += f;
if(ret == mf) return ret;
}
}
return ret;
}
int dinic(int s, int t)
{
int ans = 0;
while(bfs(s,t)) {
for(int i = 0; i <= n; i++) cur[i] = h[i];//注意要把所有的赋值
ans += dfs(s,t,INF);
}
return ans;
}
vector<int> g[maxn];
map<P,int> mp;
bool inq[maxn];
int d[maxn];
void spfa()
{
memset(d,INF,sizeof(d));
memset(inq,false,sizeof(inq));
queue<int> Q;
Q.push(1);
d[1] = 0;
inq[1] = true;
while(!Q.empty())
{
int u = Q.front();
Q.pop();
inq[u] = false;
if(d[u] > d[n]) return ;
for(int i = 0; i < g[u].size(); i++)
{
int v = g[u][i];
if(d[v] > d[u] + 1)
{
d[v] = d[u]+1;
if(!inq[v]) {inq[v] = true;Q.push(v);}
}
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
tot = 0;
for(int i = 0; i <= n; i++) g[i].clear();
mp.clear();
memset(h,-1,sizeof(h));
int u,v,c;
for(int i = 0; i < m; i++)
{
scanf("%d%d%d",&u,&v,&c);
g[u].push_back(v);
g[v].push_back(u);
mp[P(max(u,v),min(u,v))] = c;
}
spfa();
for(int i = 1; i <= n; i++)
for(int j = 0; j < g[i].size(); j++)
if(d[g[i][j]]-d[i] == 1)
{
add_edge(i,g[i][j],mp[P(max(g[i][j],i),min(g[i][j],i))]);
add_edge(g[i][j],i,0);
//cout << i << " " << g[i][j] << endl;
}
int ans = dinic(1,n);
printf("%d\n",ans);
}
return 0;
}