Til the Cows Come Home-Poj2387
题目大意:有N个点,给出从a点到b点的距离,当然a和b是互相可以抵达的,问从1到n的最短距离
思路:求最短路模板题,注意本题有重边的情况。
ac代码(bellman-ford):
//#include<bits/stdc++.h>
#include <iostream>
#include <vector>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 10;
//邻接表存图
struct Node {
int u, v, w;
};
std::vector<Node> g;
int dp[MAXN];
int t, n;
//查找s到所有点的最短路;
bool Bellman_ford(int s, int n)
{
for (int i = 0; i < n; i++) dp[i] = INF;
dp[s] = 0;
for (int k = 1; k < n; k++) {
for (int i = 0; i < g.size(); i++) {
int u = g[i].u;
int v = g[i].v;
int w = g[i].w;
dp[v] = min(dp[v], dp[u] + w);
}
}
for (int i = 0; i < g.size(); i++) {//判断是否有负环,本题不用;
int u = g[i].u;
int v = g[i].v;
int w = g[i].w;
if (dp[v] > dp[u] + w) return false;
}
return true;
}
int main()
{
while (cin >> t >> n) {
g.clear();
for (int i = 0; i < t; i++) {//存图
int u, v, w;
cin >> u >> v >> w;
g.push_back(Node{ u, v, w });
g.push_back(Node{ v, u, w });
}
Bellman_ford(n, n);
cout << dp[1] << endl;
}
return 0;
}
ac代码(dijkstar):
//#include<bits/stdc++.h>
#include<iostream>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=1000+5;
int mp[MAXN][MAXN];
int m,n;
int use[MAXN];
int dis[MAXN];
void dij(int s,int n){
int pos=-1;
dis[s]=0;
for(int i=1;i<=n;i++){
use[i]=0;
dis[i]=mp[1][i];
}
for(int i=1;i<=n;i++){
int min1=INF;
for(int j=1;j<=n;j++){
if(!use[j]&&dis[j]<min1){
pos=j;
min1=dis[j];
}
}
use[pos]=1;
if(pos==n) break;
for(int j=1;j<=n;j++){
if(!use[j]&&dis[j]>mp[pos][j]+dis[pos]){
dis[j]=mp[pos][j]+dis[pos];
}
}
}
}
int main(){
cin>>m>>n;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
mp[i][j]=INF;
}
}
for(int i=0;i<m;i++){
int u,v,w;
cin>>u>>v>>w;
if(mp[u][v]>w)
mp[u][v]=mp[v][u]=w;
}
dij(1,n);
cout<<dis[n]<<endl;
return 0;
}
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