题面
分析
考虑网络流
注意到数据包走的是最短路，所以我们只需要考虑在最短路上的边
由于最短路可能有多条，我们先跑一遍Dijkstra,然后再\(O(m)\) 遍历每条边(u,v,w)
如果dist[u]=dist[v]+w,则这条边肯定在最短路上
然后点的容量限制可以用拆点来解(常见套路)，从u向u+n连边，容量为c[u]
原图中的边(u,v)在新图中变成边(u+n,v)
然后Dinic求最大流即可
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define maxn 1005
#define maxm 200005
#define INF 0x3f3f3f3f3f3f3f3f
using namespace std;
int n,m;
vector<pair<int,long long> >G[maxn];
struct node{
int x;
long long d;
node(){
}
node(int u,long long dis){
x=u;
d=dis;
}
friend bool operator < (node p,node q){
return p.d>q.d;
}
};
long long dist[maxn];
int used[maxn];
void dijkstra(){
priority_queue<node>q;
q.push(node(1,0));
memset(dist,0x3f,sizeof(dist));
dist[1]=0;
while(!q.empty()){
node now=q.top();
q.pop();
int x=now.x;
if(used[x]) continue;
used[x]=1;
int l=G[x].size();
for(int i=0;i<l;i++){
int y=G[x][i].first;
int len=G[x][i].second;
if(dist[y]>dist[x]+len){
dist[y]=dist[x]+len;
q.push(node(y,dist[y]));
}
}
}
}
long long c[maxn];
struct edge{
int from;
int to;
int next;
long long flow;
}E[maxm<<1];
int sz=1;
int head[maxn];
void add_edge(int u,int v,long long w){
// printf("%d->%d : %d\n",u,v,w);
sz++;
E[sz].from=u;
E[sz].to=v;
E[sz].next=head[u];
E[sz].flow=w;
head[u]=sz;
sz++;
E[sz].from=v;
E[sz].to=u;
E[sz].next=head[v];
E[sz].flow=0;
head[v]=sz;
}
int deep[maxn];
bool bfs(int s,int t){
queue<int>q;
memset(deep,0,sizeof(deep));
q.push(s);
deep[s]=1;
while(!q.empty()){
int x=q.front();
q.pop();
for(int i=head[x];i;i=E[i].next){
int y=E[i].to;
if(E[i].flow&&!deep[y]){
deep[y]=deep[x]+1;
if(y==t) return 1;
q.push(y);
}
}
}
return 0;
}
long long dfs(int x,int t,long long minf){
if(x==t) return minf;
long long k,rest=minf;
for(int i=head[x];i;i=E[i].next){
int y=E[i].to;
if(E[i].flow&&deep[y]==deep[x]+1){
k=dfs(y,t,min(rest,E[i].flow));
if(k==0) deep[y]=0;
E[i].flow-=k;
E[i^1].flow+=k;
rest-=k;
}
}
return minf-rest;
}
long long dinic(int s,int t){
long long maxflow=0,nowflow=0;
while(bfs(s,t)){
while(nowflow=dfs(s,t,INF)) maxflow+=nowflow;
}
return maxflow;
}
int main(){
int u,v;
long long w;
scanf("%d %d",&n,&m);
for(int i=1;i<=m;i++){
scanf("%d %d %lld",&u,&v,&w);
G[u].push_back(make_pair(v,w));
G[v].push_back(make_pair(u,w));
}
for(int i=1;i<=n;i++) scanf("%lld",&c[i]);
c[1]=c[n]=INF;
dijkstra();
for(int i=1;i<=n;i++){
add_edge(i,i+n,c[i]);
}
for(int i=1;i<=n;i++){
for(int j=0;j<G[i].size();j++){
int y=G[i][j].first;
long long len=G[i][j].second;
if(dist[y]==dist[i]+len){
add_edge(i+n,y,INF);
}
}
}
printf("%lld\n",dinic(1,n*2));
}