题目链接:点我~~
题意:n个点m条边的有向图,f个可选城市,每个城市都有其价值w,国王的城市在1,要分配一些城市,分配的原则是:只能在规定的f个城市中选若干个,这f个城市每个都有一个价值,被选择的城市要与城市1隔离,所以需要花费一些费用来破坏边。问能获取到的最大利益,以及要破坏的边。
思路:求能获取的最大利益,先建立一个超级汇点,将f个可选城市与汇点连接,价值为权值,f个城市的价值总和减去跑出的最小割即为可获得的最大价值,过程肯定优先选取价值低的边来破坏。在残留网络中,从源点遍历所有能走到的点,即属于S集,其余的为T集,如果一条边连接着S集的点与T集的点,说明该边一定为割边。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PI;
typedef pair< PI, int> PII;
const double eps=1e-5;
const double pi=acos(-1.0);
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
#define Key_value ch[ch[root][1]][0]
#define lson l,mid,rt<<1
const int MAXN = 10000+100;
#define rson mid+1,r,rt<<1|1
const int MAXM = 200000+100;
struct Edge
{
int to,next,cap,flow;
} edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw=0)
{
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].next = head[u];
edge[tol].flow = 0;
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol].next = head[v];
edge[tol].flow = 0;
head[v] = tol++;
}
int sap(int start,int end,int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
int ans = 0;
while(dep[start] < N)
{
if(u == end)
{
int Min = INF;
for(int i = pre[u]; i != -1; i = pre[edge[i^1].to]) if(Min > edge[i].cap - edge[i].flow) Min = edge[i].cap - edge[i].flow;
for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for(int i = cur[u]; i != -1; i = edge[i].next)
{
v = edge[i].to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
{
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if(flag)
{
u = v;
continue;
}
int Min = N;
for(int i = head[u]; i != -1; i = edge[i].next)
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
{
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if(!gap[dep[u]])return ans;
dep[u] = Min+1;
gap[dep[u]]++;
if(u != start) u = edge[pre[u]^1].to;
}
return ans;
}
int S[MAXM],T[MAXM];
bool vis[MAXM];
void dfs(int u)
{
vis[u]=1;
for(int i=head[u]; ~i; i=edge[i].next)
if(!vis[edge[i].to] &&edge[i].cap>edge[i].flow )
dfs(edge[i].to);
}
int main()
{
int n,m,f;
int t,casee=1;
scanf("%d",&t);
while(t--)
{
init();
scanf("%d%d%d",&n,&m,&f);
int u,v,w;
for(int i=1; i<=m; ++i)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
S[i]=u,T[i]=v;
}
int res=0;
for(int i=0; i<f; ++i)
{
scanf("%d%d",&u,&w);
addedge(u,n+1,w);
res+=w;
}
printf("Case %d: %d\n",casee++,res-sap(1,n+1,n+1));
memset(vis,0,sizeof(vis));
dfs(1);
vector<int >ans;
for(int i=1; i<=m; ++i)
{
if(vis[S[i]] && !vis[T[i]])
ans.push_back(i);
}
cout<<ans.size();
for(auto &i:ans)
cout<<" "<<i;
puts("");
}
return 0;
}
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