首先，从1和n跑一次dij，判断每一条边能否出现在最短路上，不能出现就删掉，然后将所有边建在图上，流量为边权，跑最小割即可。
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 10005 4 #define ll long long 5 struct ji{ 6 int nex,to,len; 7 }edge[N<<1]; 8 vector<ji>v; 9 queue<int>q; 10 int E,t,n,m,x,y,z,head[N],d[N],work[N]; 11 ll d1[N],d2[N]; 12 struct ji2{ 13 int k; 14 bool operator < (const ji2 &a)const{ 15 return d1[k]>d1[a.k]; 16 } 17 }; 18 priority_queue<ji2>qq; 19 void add(int x,int y,int z){ 20 edge[E].nex=head[x]; 21 edge[E].to=y; 22 edge[E].len=z; 23 head[x]=E++; 24 } 25 void dij(int k){ 26 memset(d1,0x3f,sizeof(d1)); 27 qq.push(ji2{k}); 28 d1[k]=0; 29 while (!qq.empty()){ 30 k=qq.top().k; 31 qq.pop(); 32 for(int i=head[k];i!=-1;i=edge[i].nex){ 33 if (d1[k]+edge[i].len<d1[edge[i].to]){ 34 d1[edge[i].to]=d1[k]+edge[i].len; 35 qq.push(ji2{edge[i].to}); 36 } 37 } 38 } 39 } 40 bool bfs(){ 41 memset(d,-1,sizeof(d)); 42 q.push(1); 43 d[1]=0; 44 while (!q.empty()){ 45 int k=q.front(); 46 q.pop(); 47 for(int i=head[k];i!=-1;i=edge[i].nex){ 48 int v=edge[i].to; 49 if ((edge[i].len)&&(d[v]<0)){ 50 d[v]=d[k]+1; 51 q.push(v); 52 } 53 } 54 } 55 return d[n]>=0; 56 } 57 int dfs(int k,int s){ 58 if (k==n)return s; 59 int p; 60 for(int i=work[k];i!=-1;i=edge[i].nex){ 61 int v=edge[i].to; 62 if ((edge[i].len)&&(d[v]==d[k]+1)&&(p=dfs(v,min(s,edge[i].len)))){ 63 edge[i].len-=p; 64 edge[i^1].len+=p; 65 work[k]=i; 66 return p; 67 } 68 } 69 work[k]=-1; 70 return 0; 71 } 72 ll dinic(){ 73 int k; 74 ll ans=0; 75 while (bfs()){ 76 memcpy(work,head,sizeof(head)); 77 while (k=dfs(1,0x3f3f3f3f))ans+=k; 78 } 79 return ans; 80 } 81 int main(){ 82 scanf("%d",&t); 83 while (t--){ 84 scanf("%d%d",&n,&m); 85 if (n==1){ 86 printf("0\n"); 87 continue; 88 } 89 E=0; 90 memset(head,-1,sizeof(head)); 91 v.clear(); 92 for(int i=1;i<=m;i++){ 93 scanf("%d%d%d",&x,&y,&z); 94 add(x,y,z); 95 v.push_back(ji{x,y,z}); 96 } 97 dij(1); 98 memcpy(d2,d1,sizeof(d1)); 99 E=0; 100 memset(head,-1,sizeof(head)); 101 for(int i=0;i<v.size();i++)add(v[i].to,v[i].nex,v[i].len); 102 dij(n); 103 v.clear(); 104 for(int i=1;i<=n;i++) 105 for(int j=head[i];j!=-1;j=edge[j].nex) 106 if (d1[i]+d2[edge[j].to]+edge[j].len==d1[1])v.push_back(ji{edge[j].to,i,edge[j].len}); 107 E=0; 108 memset(head,-1,sizeof(head)); 109 for(int i=0;i<v.size();i++){ 110 add(v[i].nex,v[i].to,v[i].len); 111 add(v[i].to,v[i].nex,0); 112 } 113 printf("%lld\n",dinic()); 114 } 115 }