题意:所有的流量都只能最短路径上的边,每个点有个最大的吞吐量,求1号点到n号点的最大吞吐量
题解:嗯嗯嗯 题意就是题解啊
先求最短路,然后可以是最短路上的边连容量inf的边
每个点都拆点,然后除了1和n两个点,都,边就是他的吞吐量,1和n边当然是inf
唯一注意就是long long
1 #include<bits/stdc++.h> 2 #define pa pair<int,int> 3 #define lld long long 4 using namespace std; 5 #define N 1005 6 #define M 260000 7 #define inf 100000000000000LL 8 namespace Dinic 9 { 10 int head[N],head2[N],p=1; 11 struct Rec 12 { 13 int go,nex; 14 lld c; 15 }eg[M*2],e[M*2]; 16 void build(int a,int b,lld c) 17 { 18 eg[++p]=(Rec){b,head[a],-c}; 19 head[a]=p; 20 eg[++p]=(Rec){a,head[b],0}; 21 head[b]=p; 22 } 23 lld dis[N],ans; 24 int Q[N],s[N],S,T,stop; 25 bool bfs() 26 { 27 memset(dis,0,sizeof(dis)); 28 dis[T]=1; 29 Q[1]=T; 30 for (int p1=1,p2=1;p1<=p2;p1++) 31 { 32 for (int i=head[Q[p1]];i;i=eg[i].nex) 33 if (eg[i^1].c<0&&!dis[eg[i].go]) 34 { 35 dis[eg[i].go]=dis[Q[p1]]+1; 36 Q[++p2]=eg[i].go; 37 } 38 } 39 if (!dis[S]) return false; 40 memcpy(head2,head,sizeof(head)); 41 return true; 42 } 43 bool dinic(int p,int top) 44 { 45 if (p==T) 46 { 47 lld x=inf; 48 for (int i=1;i<=top-1;i++) if (-eg[s[i]].c<x) x=-eg[s[i]].c,stop=i; 49 for (int i=1;i<=top-1;i++) eg[s[i]].c+=x,eg[s[i]^1].c-=x; 50 ans+=x; 51 return true; 52 } 53 for (int &i=head2[p];i;i=eg[i].nex) 54 { 55 if (eg[i].c<0&&dis[eg[i].go]==dis[p]-1) 56 { 57 s[top]=i; 58 if (dinic(eg[i].go,top+1)&&top!=stop) return true; 59 } 60 } 61 return false; 62 } 63 lld ask() 64 { 65 ans=0; 66 while (bfs()) dinic(S,1); 67 return ans; 68 } 69 void init(int _S,int _T){ 70 S=_S,T=_T; 71 } 72 } 73 using namespace Dinic; 74 void clear() 75 { 76 p=1; 77 memset(head,0,sizeof(head)); 78 } 79 int n,m,ss,tt,x[M],y[M]; 80 long long xx,diss[N],z[M]; 81 void addedge(int a,int b,lld c) 82 { 83 p++; 84 e[p].go=b; 85 e[p].c=c; 86 e[p].nex=head[a]; 87 head[a]=p; 88 } 89 void dijkstra() 90 { 91 priority_queue<pa,vector<pa>,greater<pa> >q; 92 for (int i=1;i<=n;i++) diss[i]=inf; 93 diss[1]=0; 94 q.push(make_pair(0,1)); 95 while (!q.empty()) 96 { 97 int now=q.top().second; 98 q.pop(); 99 for (int i=head[now];i;i=e[i].nex) 100 if(diss[now]+e[i].c<diss[e[i].go]) 101 { 102 diss[e[i].go]=diss[now]+e[i].c; 103 q.push(make_pair(diss[e[i].go],e[i].go)); 104 } 105 } 106 } 107 int main() 108 { 109 scanf("%d%d",&n,&m); 110 ss=1;tt=n+n; 111 init(ss,tt); 112 for (int i=1;i<=m;i++) 113 { 114 scanf("%d%d%lld",&x[i],&y[i],&z[i]); 115 addedge(x[i],y[i],z[i]); 116 addedge(y[i],x[i],z[i]); 117 } 118 dijkstra(); 119 clear(); 120 for (int i=1;i<=m;i++) 121 { 122 if (diss[x[i]]+z[i]==diss[y[i]]) build(x[i]+n,y[i],inf); 123 if (diss[y[i]]+z[i]==diss[x[i]]) build(y[i]+n,x[i],inf); 124 } 125 for (int i=1;i<=n;i++) 126 { 127 scanf("%lld",&xx); 128 if (i==1 || i==n) build(i,i+n,inf);else build(i,i+n,xx); 129 } 130 printf("%lld",ask()); 131 return 0; 132 }