正题

题目链接:https://www.luogu.com.cn/problem/P4300

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题目大意

$n$个点$m$条边的无向图。求$1\sim n$的最短路和删除$c_i$和最小的边使得最短路变长。

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解题思路

显然我们需要跑一次最短路。

之后考虑如何求第二问，我们发现我们要割掉最短路上的边，所以我们只要把最短路树（一张有最短路上的边构成的$DAG$）构出来然后在上面求最小割就好了。

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N=510,inf=2147483647/3;
struct edge_node{int x,y,w,c;
}e[124760];
struct point{int pos,val;
};
struct node{int to,next,w;
}a[124760*2];
bool operator<(point x,point y)
{return x.val>y.val;}
int n,m,s,t,tot,f[N],ls[N],dep[N],ans;
bool v[N];queue<int> qq;
priority_queue<point> q;
void addl(int x,int y,int w){a[++tot].to=y;a[tot].next=ls[x];ls[x]=tot;a[tot].w=w;return;
}
void dij(){memset(f,0x3f,sizeof(f));q.push((point){1,0});f[1]=0;while(!q.empty()){int x=q.top().pos;q.pop();if(v[x])continue;v[x]=1;for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(f[x]+a[i].w<f[y]){f[y]=f[x]+a[i].w;if(!v[y])q.push((point){y,f[y]});}}}return;
}
bool bfs(){while(!qq.empty())qq.pop();qq.push(1);memset(dep,0,sizeof(dep));dep[s]=s;while(!qq.empty()){int x=qq.front();qq.pop();for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(dep[y]||!a[i].w)continue;dep[y]=dep[x]+1;if(y==t)return 1;qq.push(y);}}return 0;
}
int dinic(int x,int flow){if(x==t)return flow;int rest=0,k;for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(dep[x]+1!=dep[y]||!a[i].w)continue;rest+=(k=dinic(y,min(a[i].w,flow-rest)));a[i].w-=k;a[i^1].w+=k;if(flow==rest)return flow;}if(!rest)dep[x]=0;return rest;
}
void dfs(int x,int fa){v[x]=1;for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(f[y]+a[i].w==f[x])if(!v[y])dfs(y,x);}
}
int main()
{scanf("%d%d",&n,&m);s=1;t=n;for(int i=1;i<=m;i++){scanf("%d%d%d%d",&e[i].x,&e[i].y,&e[i].w,&e[i].c);addl(e[i].x,e[i].y,e[i].w);addl(e[i].y,e[i].x,e[i].w);}dij();tot=1;memset(v,0,sizeof(v));dfs(n,n);memset(ls,0,sizeof(ls));for(int i=1;i<=m;i++){if(!v[e[i].x]||!v[e[i].y])continue;if(f[e[i].x]+e[i].w==f[e[i].y]){addl(e[i].x,e[i].y,e[i].c);addl(e[i].y,e[i].x,0);}if(f[e[i].y]+e[i].w==f[e[i].x]){addl(e[i].x,e[i].y,0);addl(e[i].y,e[i].x,e[i].c);}}while(bfs())ans+=dinic(s,inf);printf("%d\n%d\n",f[n],ans);
}