两道最短路好题

POJ3037

手玩一下 发现每一点的速度可以直接搞出来，就是pow(2,h[1][1]-h[i][j])*V

那么从这个点出发到达别的点的耗费的时间都是上面这个数的倒数，然后直接跑最短路就好了

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;

const int N = 1e5+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int gcd(int a,int b){return b?a:gcd(b,a%b);}
int lcm(int a,int b){return a*b/gcd(a,b);}
int qmi(int a,int b,int mod){int res=1;while(b){if(b&1)res=res*a%mod;b>>=1;a=a*a%mod;}return res;}


int n,q,m,v;
bool vis[1010][1010];
// int e[N],ne[N],w[N],h[N],idx;
// void add(int a,int b,int c){
	// e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
// }

double vs[1010][1010];
double dist[1010][1010];
double h[1010][1010];



void solve()
{
	cin>>v>>n>>m;
	
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			double x;cin>>x;
			h[i][j] = x;
			vs[i][j] = pow(2,x-h[1][1])/v; 
			dist[i][j] = 1e15;
		}
	}
	
	dist[1][1] = 0;
	queue<pair<int,int>>q;
	q.push(make_pair(1,1));
	int dx[] = {0,0,1,-1};
	int dy[] = {1,-1,0,0};
	vis[1][1] = true;
	
	while(q.size()){
		pair<int,int> t = q.front();
		q.pop();
		int x = t.first,y = t.second;
		vis[x][y] = false;
		//cout<<x<<" "<<y<<"\n";
		
		for(int i=0;i<4;i++){
			int temx = x+dx[i],temy = y+dy[i];
			if(temx<1||temx>n||temy<1||temy>m)continue;
			//cout<<temx<<" "<<temy<<"\n";
			if(dist[temx][temy]>dist[x][y]+vs[x][y]){
				dist[temx][temy] = dist[x][y]+vs[x][y];
				if(!vis[temx][temy]){
					vis[temx][temy] = true;
					q.push(make_pair(temx,temy));
				}
			}
		}
		
	}
	
	printf("%.2lf",dist[n][m]);
	
	

}

signed main()
{
	//ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int _;
	//cin>>_;
	_ = 1;
	while(_--)solve();
	return 0;
}

HDU6714

这个dijk记数还是很有意思的，你得明白folyd的含义但是别被DP的含义绕进去

每次暴力的跑每一个点的单源最短路，然后当有中间点的时候你就更新一下就行了，没有中间的时候D【i】【j】就是一开始的距离，没有被更新，还是很有趣的，还是得想明白floyd的具体过程（好像不懂也行

一开始我就被绕进去了，一直在扣floyd 的含义来写这道，发现直接按上面的做法就好了

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define int long long
using pii = pair<int,int>;
const int N = 1e4+10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
int gcd(int a,int b){return b?a:gcd(b,a%b);}
int lcm(int a,int b){return a*b/gcd(a,b);}
int qmi(int a,int b,int mod){int res=1;while(b){if(b&1)res=res*a%mod;b>>=1;a=a*a%mod;}return res;}


int n,q,m;
int id[N];
int e[N],ne[N],w[N],h[N],idx;
void add(int a,int b,int c){
	e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
}

bool vis[N];
ll dist[N];



void dijkstra(int mid)
{
	memset(dist,0x3f,sizeof dist);
	memset(vis,0,sizeof vis);
	memset(id,0,sizeof id);
	priority_queue<pii,vector<pii>,greater<pii>>heap;
	heap.push({0,mid});
	dist[mid] = 0;
	
	
	while(heap.size()){
		auto t = heap.top();
		heap.pop();
		int ver = t.second;
		
		if(vis[ver])continue;
		vis[ver] = true;
		//cout<<ver<<"\n";
		
		for(int i=h[ver];~i;i=ne[i]){
			int j = e[i];
			if(dist[j]>dist[ver]+w[i]){
				dist[j] = dist[ver]+w[i];
				heap.push({dist[j],j});
				if(ver==mid)continue;
				id[j] = max(id[ver],ver);
				
			}else if(dist[j]==dist[ver]+w[i]){
				id[j] = min(id[j],max(id[ver],ver));
			}
		}
		
	}
}


void solve()
{
	cin>>n>>m;
	memset(h,-1,sizeof h);
	idx = 0;
	while(m--){
		int a,b,c;cin>>a>>b>>c;
		add(a,b,c),add(b,a,c);
	}
	
	int ans = 0;
	for(int i=1;i<=n;i++){
		dijkstra(i);
		for(int j=1;j<=n;j++)ans = (id[j]+ans)%mod;
		//cout<<"\n";
	
	}
	
	cout<<ans;

}

signed main()
{
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int _;
	cin>>_;
	//_ = 1;
	while(_--)solve();
	return 0;
}