codeforces1433G Reducing Delivery Cost

https://codeforces.com/contest/1433/problem/G

998ms卡过去了。。。

n^3floyed跑出来每个点到另一个点的答案

然后枚举删除每条边,再枚举每个route,可以O(1)找到此时这个route的最小代价,加起来得到删掉这条边的答案,复杂度为O(mk)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxl=1010;
const int inf=1e9+10;

int n,m,k,cnt,tot,cas,ans;
int f[maxl][maxl];
struct edge
{
	int u,v,w;
}edg[maxl*maxl];
struct ed
{
	int v,l;
};
struct route
{
	int a,b;
}a[maxl];
vector<ed> e[maxl];
bool vis[maxl][maxl];
char s[maxl];

inline void prework()
{
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			f[i][j]=inf;
	for(int i=1;i<=n;i++)
		f[i][i]=0;
	int u,v,w;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&u,&v,&w);
		edg[i]=edge{u,v,w};
		f[u][v]=f[v][u]=min(f[u][v],w);
	}
	for(register int k=1;k<=n;++k)
		for(register int i=1;i<=n;++i)
		if(k!=i)
			for(register int j=1;j<=n;++j)
			if(j!=i && k!=j)
				f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
	for(int i=1;i<=k;i++)
		scanf("%d%d",&a[i].a,&a[i].b);	
}

inline void mainwork()
{
	int u,v,tmp;ans=inf;
	for(register int i=1;i<=m;++i)
	{
		u=edg[i].u;v=edg[i].v;
		tmp=0;
		for(register int j=1;j<=k;++j)
			tmp+=min(f[a[j].a][a[j].b],min(f[a[j].a][u]+f[a[j].b][v],f[a[j].a][v]+f[a[j].b][u]));
		ans=min(ans,tmp);
	}
}

inline void print()
{
	printf("%d",ans);
}

int main()
{
	int t=1;
	//scanf("%d",&t);
	for(cas=1;cas<=t;cas++)
	{
		prework();
		mainwork();
		print();
	}
	return 0;
}

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转载自blog.csdn.net/liufengwei1/article/details/109192600