Meaning of the questions:
Give you a directed graph, and now you need to delete some edges, each deletion cost is edge weights, so that the shortest path is increased, and now ask your minimum cost.
answer:
If you want to make the most short-circuit increases, apparently deleted some of the shortest side of the road. We found that most of the original short circuit can be formed a \ (\ text the DAG} {\) , so that if we want to increase the shortest, is equivalent to a required minimum cost, so that we can destroy the \ (\ text DAG} {\) , but apparently this thing is the minimum cut. So we just need to build the original image of the shortest side, cut to a minimum after running on the new map. As for the original construction shortest path side, we only need to \ (1 \) number of nodes, and \ (\ n-) numbers are for the most short-circuit node, then two nodes for each edge connected \ (U \) and \ (V \) , if the \ (dis_ {1u} + dis_ {nv} + val_ {uv} = dis_ {1n} \) is the shortest proof strip edge is the edge of the road.
Looked at the data, it looks as though less powerful ah, in order not to T, the amount of data in such a good best or starting a maximum flow algorithm ......
Code:
#include <bits/stdc++.h>
#define maxn 30005
using namespace std;
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
int head1[maxn],cnt1;
int head2[maxn],cnt2;
int n,m;
struct edge{
int to,next,from;
long long cost;
}q1[maxn],q2[maxn];
void add_edge(int from,int to,int cost,int *head,int &cnt,edge *q){
q[cnt].to=to;
q[cnt].from=from;
q[cnt].cost=cost;
q[cnt].next=head[from];
head[from]=cnt++;
}
typedef pair<int,ll>P;
ll d1[maxn],d2[maxn],e[maxn],h[maxn],cnth[maxn];
void diji(int s,int *head,edge *q,ll *d){
for(int i=1;i<=n;i++) d[i]=INF;
priority_queue<P,vector<P>,greater<P> >que;
d[s]=0;
que.push(P(0,s));
while(!que.empty()){
P p=que.top();
que.pop();
int x=p.second;
if(d[x]<p.first) continue;
for(int i=head[x];i!=-1;i=q[i].next){
edge id=q[i];
if(d[id.to]>d[x]+id.cost){
d[id.to]=d[x]+id.cost;
que.push((P(d[id.to],id.to)));
}
}
}
}
//hlpp求解最大流
int head3[maxn],cntt=2;
struct Node{
int to,next;
ll val;
}qq[maxn<<1];
int vis[maxn];
int sp,ep;
struct cmp{
inline bool operator()(int a,int b) const{
return h[a]<h[b];
}
};
void addedge(int from,int to,int val){
qq[cntt].to=to;
qq[cntt].val=val;
qq[cntt].next=head3[from];
head3[from]=cntt++;
}
void add_edge2(int from,int to,int val){
addedge(from,to,val);
addedge(to,from,0);
}
void bfs(){
memset(h,0x3f,sizeof(h));
h[ep]=0;
queue<int>que;
que.push(ep);
while(!que.empty()){
int x=que.front();
que.pop();
vis[x]=0;
for(int i=head3[x];i!=-1;i=qq[i].next){
int to=qq[i].to;
if(qq[i^1].val&&h[to]>h[x]+1){
h[to]=h[x]+1;
if(vis[to]==0){
que.push(to);
vis[to]=1;
}
}
}
}
return;
}
void init(int *head,int &cnt){
for(int i=0;i<maxn;i++) head[i]=-1;
memset(vis,0,sizeof(vis));
memset(cnth,0,sizeof(cnth));
memset(e,0,sizeof(e));
cnt=0;
}
priority_queue<int,vector<int>,cmp>Q;
inline void push_(int x){
for(int i=head3[x];i!=-1;i=qq[i].next){
int to=qq[i].to;
if(qq[i].val&&h[to]+1==h[x]){
int mi=min(qq[i].val,e[x]);
qq[i].val-=mi;
qq[i^1].val+=mi;
e[x]-=mi;
e[to]+=mi;
if(vis[to]==0&&to!=ep&&to!=sp){
Q.push(to);
vis[to]=1;
}
if(e[x]==0)break;
}
}
}
inline void relabel(int x){
h[x]=INF;
for(int i=head3[x];i!=-1;i=qq[i].next){
int to=qq[i].to;
if(qq[i].val&&h[to]+1<h[x]){
h[x]=h[to]+1;
}
}
}
ll hlpp(){
register int i;
bfs();
if(h[sp]==INF)return 0;
h[sp]=n;
for(i=1;i<=n;i++)if(h[i]<INF)cnth[h[i]]++;
for(i=head3[sp];i!=-1;i=qq[i].next){
int to=qq[i].to;
ll mi=qq[i].val;
if(mi){
e[sp]-=mi;
e[to]+=mi;
qq[i].val-=mi;
qq[i^1].val+=mi;
if(to!=ep&&vis[to]==0&&to!=sp){
Q.push(to);
vis[to]=1;
}
}
}
while(!Q.empty()){
int x=Q.top();
vis[x]=0;
Q.pop();
push_(x);
if(e[x]){
cnth[h[x]]--;
if(cnth[h[x]]==0){
for(int i=1;i<=n;i++){
if(i!=sp&&i!=ep&&h[i]>h[x]&&h[i]<n+1){
h[i]=n+1;
}
}
}
relabel(x);
cnth[h[x]]++;
Q.push(x);
vis[x]=1;
}
}
return e[ep];
}
int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
init(head1,cnt1);
init(head2,cnt2);
init(head3,cntt);
cntt=2;
for(int i=1;i<=m;i++){
int from,to,cost;
scanf("%d%d%d",&from,&to,&cost);
add_edge(from,to,cost,head1,cnt1,q1);
add_edge(to,from,cost,head2,cnt2,q2);
}
diji(1,head1,q1,d1);
diji(n,head2,q2,d2);
if(d1[n]==INF){
puts("0");
continue;
}
sp=1,ep=n;
for(int i=1;i<=n;i++){
for(int j=head1[i];j!=-1;j=q1[j].next){
int to=q1[j].to;
int from=q1[j].from;
int val=q1[j].cost;
if(d1[from]+d2[to]==d1[n]-val){
add_edge2(from,to,val);
}
}
}
printf("%lld\n",hlpp());
}
return 0;
}