BZOJ 3931 (网络流+最短路）

题面

传送门

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));
}