Barricade
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1117 Accepted Submission(s): 340
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.
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
4 4
1 2 1
2 4 2
3 1 3
4 3 4
Sample Output
4
最短路+网络流。
先一遍bfs找到最短路，再一次bfs找到最短路上的点，通过dis[i]+1 = dis[u]来找，然后跑一遍Dinic。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int maxn = ;
const int inf = 0x3f3f3f3f;
int n,m;
int g[maxn][maxn];
int vis[maxn];
int dis[maxn];
struct edge
{
int to;
int cap;
int rev;
};
vector<edge> gg[maxn];
int level[maxn];
int it[maxn];
void add(int from,int to,int cap)
{
edge cur;
cur.to = to;
cur.cap = cap;
cur.rev = gg[to].size();
gg[from].push_back(cur);
cur.to = from;
cur.cap = ;
cur.rev = gg[from].size()-;
gg[to].push_back(cur);
} void bfs(int s)
{
memset(level,-,sizeof(level));
queue<int> q;
level[s] = ;
q.push(s);
while(!q.empty())
{
int v = q.front(); q.pop();
for(int i=;i<gg[v].size();i++)
{
edge &e = gg[v][i];
if(e.cap>&&level[e.to]<)
{
level[e.to] = level[v]+;
q.push(e.to);
}
}
}
}
int dfs(int v,int t,int f)
{
if(v==t) return f;
for(int &i=it[v];i<gg[v].size();i++)
{
edge &e = gg[v][i];
if(e.cap>&&level[v]<level[e.to])
{
int d = dfs(e.to,t,min(f,e.cap));
if(d>)
{
e.cap -= d;
gg[e.to][e.rev].cap += d;
return d;
}
}
}
return ;
}
int max_flow(int s,int t)
{
int flow = ;
for(;;)
{
bfs(s);
if(level[t]<) return flow;
memset(it,,sizeof(it));
int f;
while((f=dfs(s,t,inf))>) flow += f;
}
}
bool bfs1()
{
queue<int> q;
memset(vis,,sizeof(vis));
memset(dis,inf,sizeof(dis));
vis[] = ;
dis[] = ;
q.push();
while(!q.empty())
{
int cur = q.front();q.pop();
if(cur==n) return true;
for(int i=;i<=n;i++)
{
if(cur==i) continue;
if(!vis[i]&&g[cur][i]!=-)
{
vis[i] = ;
dis[i] = dis[cur]+;
q.push(i);
}
}
}
return false;
}
void bfs2()
{
queue<int> q;
memset(vis,,sizeof(vis));
vis[n] = ;
q.push(n);
while(!q.empty())
{
int cur = q.front();q.pop();
for(int i=;i<=n;i++)
{
if(cur==i) continue;
if(g[cur][i]==-) continue;
if(dis[i]+==dis[cur])
{
add(i,cur,g[i][cur]);
if(!vis[i])
{
vis[i] = ;
q.push(i);
}
}
}
}
}
int main()
{
int T;cin>>T;
while(T--)
{
scanf("%d %d",&n,&m);
int u,v,w;
memset(g,-,sizeof(g));
for(int i=;i<maxn;i++) gg[i].clear();
for(int i=;i<=m;i++)
{
scanf("%d %d %d",&u,&v,&w);
g[u][v] = w;
g[v][u] = w;
}
int ans = ;
bfs1();
bfs2();
ans = max_flow(,n);
printf("%d\n",ans);
}
return ;
}