HDU2544最短路（DFS/Dijkstra+heap）

题意：求权值最小的路，题目保证一定有解。

思路：经典的最短路问题，floyd、Dijkstra或者DFS都可以。

//#include<bits/stdc++.h>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF  0x3f3f3f3f
typedef long long  LL;
int m,n,ans,g[105][105],vis[105];
void dfs(int d,int l){//当前结点、已有路径长度
if(l>=ans)return;
if(d==n){
ans=l;
return;
}
for(int i=2;i<=n;++i){
if(!vis[i]&&g[d][i]!=0){
vis[i]=1;
dfs(i,l+g[d][i]);
vis[i]=0;
}
}
}
int main(){
int a,b,c;
while(scanf("%d%d",&n,&m),m+n){
memset(vis,0,sizeof(vis));
memset(g,0,sizeof(g));
for(int i=0;i<m;++i){
scanf("%d%d%d",&a,&b,&c);
g[a][b]=g[b][a]=c;
}
ans=INF;
dfs(1,0);
printf("%d\n",ans);
}
}

练习图论再次看到这个题，用Dijkstra加上堆优化，没有什么坑，可以拿来当模版。

太久不用优先队列在重载小于卡了很久，priority_queue中定义的比较规则是决定在队列里的优先级，跟set之类是反着的，长记性。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<set>
#include<stack>
#include<queue>
#include<list>
#include<algorithm>
#include<queue>
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int N=10001;
struct ed{
int v,c,next;
}edge
;
struct node{
int w,u;
node(int u,int w):u(u),w(w){}
bool operator < (const node& b)const{
if(w==b.w)return u<b.u;
return w>b.w;
}
};
int n,cnt,dis
,vis
,pre
;
void addEdge(int u,int v,int c){//建立邻接矩阵
edge[cnt].v=v;
edge[cnt].c=c;
edge[cnt].next=pre[u];
pre[u]=cnt++;
}
void dijkstra(int s){
int cur;
priority_queue<node> q;
for(int i=1;i<=n;++i){
dis[i]=inf;
vis[i]=0;
}
dis[s]=0;
q.push(node(s,0));
while(!q.empty()){
cur=q.top().u;
q.pop();
if(vis[cur])continue;
vis[cur]=1;
for(int i=pre[cur];i!=-1;i=edge[i].next){
if(!vis[edge[i].v]&&dis[edge[i].v]>dis[cur]+edge[i].c){
dis[edge[i].v]=dis[cur]+edge[i].c;
q.push(node(edge[i].v,dis[edge[i].v]));
}
}
}
}
int main(){
int m,a,b,c;
while(~scanf("%d%d",&n,&m),m+n){
for(int i=1;i<=n;++i){
pre[i]=-1;
}
cnt=0;
while(m--){
scanf("%d%d%d",&a,&b,&c);
addEdge(a,b,c);
addEdge(b,a,c);
}
dijkstra(1);
printf("%d\n",dis
);
}
}
受到启发，用set也行。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<set>
#include<stack>
#include<queue>
#include<list>
#include<algorithm>
#include<queue>
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int N=10001;
struct ed{
int v,c,next;
}edge
;
struct node{
int w,u;
node(int u,int w):u(u),w(w){}
bool opera
b9ba
tor < (const node& b)const{
if(w==b.w)return u>b.u;
return w<b.w;
}
};
int n,cnt,dis
,vis
,pre
;
void addEdge(int u,int v,int c){
edge[cnt].v=v;
edge[cnt].c=c;
edge[cnt].next=pre[u];
pre[u]=cnt++;
}
void dijkstra(int s){
int cur;
set<node> se;
for(int i=1;i<=n;++i){
dis[i]=inf;
vis[i]=0;
}
dis[s]=0;
se.insert(node(s,0));
while(!se.empty()){
cur=se.begin()->u;
se.erase(se.begin());
if(vis[cur])continue;
vis[cur]=1;
for(int i=pre[cur];i!=-1;i=edge[i].next){
if(!vis[edge[i].v]&&dis[edge[i].v]>dis[cur]+edge[i].c){
dis[edge[i].v]=dis[cur]+edge[i].c;
se.insert(node(edge[i].v,dis[edge[i].v]));
}
}
}
}
int main(){
int m,a,b,c;
while(~scanf("%d%d",&n,&m),m+n){
for(int i=1;i<=n;++i){
pre[i]=-1;
}
cnt=0;
while(m--){
scanf("%d%d%d",&a,&b,&c);
addEdge(a,b,c);
addEdge(b,a,c);
}
dijkstra(1);
printf("%d\n",dis
);
}
}