BarricadeTime Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2942 Accepted Submission(s): 829 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.
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
Recommend wange2014
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給出N個點M條邊，敵人在N點，每次都會選擇最短路徑來攻擊1點，初始邊權爲1，想要改變第i條邊價值爲wi，問 現在想堵住一些路使得最短路徑增大，需要增加邊權的最小值。
跑一遍最短路之後，然後
for(int u = 1; u <= n; u ++)
{
for(int i = DJ.head[u]; ~i; i = DJ.nxt[i])
{
int v = DJ.to[i];
ll cst = DJ.val[i];
if(d[v]+1 == d[u])
{
DC.ade(u, DJ.to[i], cst), DC.ade(DJ.to[i],u,0);
}
}
}這樣就把最短路徑上的點存了下來。
然後再跑一遍最大流最小割，求出答案。
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn = 1e4 + 5;
typedef pair<ll, int> P;
const ll INF = 0x3f3f3f3f3f3f3f3f;
int n, m, cas;
ll d[maxn];
struct GDJ
{
int head[maxn], cnt;
int to[maxn << 1], nxt[maxn << 1];
ll val[maxn << 1];
void init()
{
memset(head, -1, (n + 5) * sizeof(int));
cnt = -1;
}
void ade(int a, int b, ll c)
{
to[++cnt] = b;
nxt[cnt] = head[a];
val[cnt] = c;
head[a] = cnt;
}
bool vis[maxn];
void dj(int s)
{
priority_queue<P, vector<P>, greater<P> > que;
memset(d, 0x3f, (n + 5) * sizeof(ll));
memset(vis, 0, (n + 5) * sizeof(bool));
d[s] = 0;
que.push(P(0, s));
while(!que.empty())
{
P p = que.top();
que.pop();
int u = p.second;
if(vis[u])
continue;
vis[u] = 1;
for(int i = head[u]; ~i; i = nxt[i])
{
int v = to[i];
if(d[v] > d[u] + 1)
{
d[v] = d[u] + 1;
que.push(P(d[v], v));
}
}
}
}
} DJ;
struct GDC
{
int depth[maxn], cur[maxn], head[maxn], cnt;
int to[maxn << 1], nxt[maxn << 1];
ll val[maxn << 1];
void init()
{
memset(head, -1, (n + 5) * sizeof(ll));
cnt = -1;
}
void ade(int a, int b, ll c)
{
to[++cnt] = b;
nxt[cnt] = head[a];
val[cnt] = c;
head[a] = cnt;
}
bool bfs()
{
queue<int> que;
que.push(1);
memset(depth, 0, (n + 5) * sizeof(int));
depth[1] = 1;
while(!que.empty())
{
int u = que.front();
que.pop();
for(int i = head[u]; ~i; i = nxt[i])
{
if(val[i] > 0 && depth[to[i]] == 0)
{
depth[to[i]] = depth[u] + 1;
que.push(to[i]);
}
}
}
if(depth[n])
return 1;
else
return 0;
}
ll dfs(int u, ll dist)
{
if(u == n)
return dist;
for(int &i = cur[u]; ~i; i = nxt[i])
{
if(depth[to[i]] == depth[u] + 1 && val[i] > 0)
{
ll tmp = dfs(to[i], min(dist, val[i]));
if(tmp > 0)
{
val[i] -= tmp;
val[i ^ 1] += tmp;
return tmp;
}
}
}
return 0;
}
ll dinic()
{
ll res =0, d;
while(bfs())
{
for(int i = 0; i <= n; i ++)
cur[i] = head[i];
while(d = dfs(1,INF))
res += d;
}
return res;
}
} DC;
int main()
{
scanf("%d",&cas);
while(cas --)
{
scanf("%d %d",&n,&m);
DJ.init();
for(int i = 1; i <= m; i++)
{
int a, b;
ll c;
scanf("%d %d %d",&a,&b,&c);
DJ.ade(a, b, c);
DJ.ade(b, a, c);
}
DJ.dj(n);
DC.init();
for(int u = 1; u <= n; u ++)
{
for(int i = DJ.head[u]; ~i; i = DJ.nxt[i])
{
int v = DJ.to[i];
ll cst = DJ.val[i];
if(d[v]+1 == d[u])
{
DC.ade(u, DJ.to[i], cst), DC.ade(DJ.to[i],u,0);
}
}
}
printf("%lld\n",DC.dinic());
}
return 0;
}