基础模板：
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1010;
int n,m,x,ct;
int g[maxn][maxn],gr[maxn][maxn],dist[maxn],vis[maxn];
struct node
{
int id,fi,gi;
friend bool operator < (node a,node b)
{
if(a.fi == b.fi)
return a.gi > b.gi;
return a.gi > b.gi;
}
}s[20000010];
int init()
{
memset(dist,0x3f,sizeof(dist));
for(int i = 0;i <= n;i ++)
vis[i] = 1;
dist[n - 1] = 0;
for(int i = 0;i < n;i ++)
{
int k = n;
for(int j = 0;j < n;j ++)
if(vis[j] && dist[j] < dist[k])
k = j;
if(k == n)
break;
vis[k] = 0;
for(int j = 0;j < n;j ++)
if(vis[j] && dist[k] + gr[k][j] < dist[j])
dist[j] = dist[k] + gr[k][j];
}
return 1;
}
int solve()
{
if(dist[0] == inf)
return -1;
ct = 0,s[ct].id = 0,s[ct].fi = 0;
s[ct++].gi = dist[0];
int cnt = 0;
while(ct)
{
int id = s[0].id;
int fi = s[0].fi;
if(id == n - 1)
cnt++;
if(cnt == x)
return fi;
pop_heap(s,s + ct);
ct--;
for(int j = 0;j < n;j ++)
{
if(g[id][j] < inf)
{
s[ct].id = j;
s[ct].fi = fi + g[id][j];
s[ct].gi = s[ct].fi + dist[j];
ct ++;
push_heap(s,s + ct);
}
}
}
return -1;
}
int main()
{
while(cin >> n >> m >> x)
{
memset(g,0x3f,sizeof(g));
memset(gr,0x3f,sizeof(gr));
for(int i = 0;i < m;i ++)
{
int p,q,r;
cin >> p >> q >> r;
p--;
q--;
g[p][q] = g[p][q] <= r ? g[p][q] : r;
gr[q][p] = gr[q][p] <= r ? gr[q][p] : r;
}
init();
cout << solve() << endl;
}
return 0;
}
例：
2 2 2
1 2 5
2 1 4
14例题1：poj2449 Remmarguts' Date
。。。。。
Input
The first line contains two integer numbers N and M (1 <= N <= 1000, 0 <= M <= 100000). Stations are numbered from 1 to N. Each of the following M lines contains three integer numbers A, B and T (1 <= A, B <= N, 1 <= T <= 100). It shows that there is a directed sideway from A-th station to B-th station with time T.
The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).
Output
A single line consisting of a single integer number: the length (time required) to welcome Princess Uyuw using the K-th shortest path. If K-th shortest path does not exist, you should output "-1" (without quotes) instead.
Sample Input
2 2
1 2 5
2 1 4
1 2 2
Sample Output
14题意：求两点间的第K短路（A*算法模板）（摘自此处）
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define inf 1<<29
#define M 500200
#define N 500020
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,m,k,s,t;
int head[N],pos;
int head1[N],pos1;
struct edge{int to,c,next;}ee[M<<1],e[M<<1];
void add(int a,int b,int c)
{pos++;e[pos].to=b,e[pos].c=c,e[pos].next=head[a],head[a]=pos;}
void add1(int b,int a,int c)
{pos1++;ee[pos1].to=b,ee[pos1].c=c,ee[pos1].next=head1[a],head1[a]=pos1;}
queue<int>Q;bool vis[N];int dis[N];
void spfa(int st)
{
for(int i=1;i<=n;i++)
dis[i]=inf,vis[i]=0;
dis[st]=0,vis[st]=1;Q.push(st);
while(!Q.empty())
{
int u=Q.front();Q.pop();vis[u]=0;
for(int i=head1[u];i;i=ee[i].next)
{
int v=ee[i].to;
if(dis[v]>dis[u]+ee[i].c)
{
dis[v]=dis[u]+ee[i].c;
if(!vis[v])Q.push(v),vis[v]=1;
}
}
}
}
struct node{int x,d,f;}u,v;
bool operator < (node a,node b)
{
if(a.f!=b.f)return a.f>b.f;
}
priority_queue<node>que;
void solve()
{
while(!que.empty())que.pop();
if(dis[s]==inf)
{
printf("-1
");
return ;
}
u.f=dis[s],u.x=s,u.d=0;
int now=0;que.push(u);
while(!que.empty())
{
u=que.top();que.pop();
if(u.x==t)
now++;
if(now==k)
{
printf("%d
",u.d);
return;
}
for(int i=head[u.x];i;i=e[i].next)
{
int tc=e[i].to;v=u;
v.x=tc,v.d+=e[i].c;
v.f=dis[v.x]+v.d;
que.push(v);
}
}printf("-1
");
}
int main()
{
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
while(~scanf("%d%d",&n,&m))
{
memset(head,0,sizeof(head));
memset(head1,0,sizeof(head1));
pos1=pos=0;
for(int i=1;i<=m;i++)
{
int x=read(),y=read();
int c=read();
add(x,y,c);
add1(x,y,c);
}s=read(),t=read(),k=read();
if(s==t)k++;
spfa(t);solve();
}
}例题2：ACM-ICPC 2018 沈阳赛区网络预赛 Made In Heaven
。。。。。。。
Input
There are at most 505050 test cases.
The first line contains two integers NNN and MMM (1≤N≤1000,0≤M≤10000)(1 leq N leq 1000, 0 leq M leq 10000)(1≤N≤1000,0≤M≤10000). Stations are numbered from 111 to NNN.
The second line contains four numbers S,E,KS, E, KS,E,K and TTT ( 1≤S,E≤N1 leq S,E leq N1≤S,E≤N, S≠ES eq ES≠E, 1≤K≤100001 leq K leq 100001≤K≤10000, 1≤T≤1000000001 leq T leq 1000000001≤T≤100000000 ).
Then MMM lines follows, each line containing three numbers U,VU, VU,V and WWW (1≤U,V≤N,1≤W≤1000)(1 leq U,V leq N, 1 leq W leq 1000)(1≤U,V≤N,1≤W≤1000) . It shows that there is a directed road from UUU-th spot to VVV-th spot with time WWW.
It is guaranteed that for any two spots there will be only one directed road from spot AAA to spot BBB (1≤A,B≤N,A≠B)(1 leq A,B leq N, A eq B)(1≤A,B≤N,A≠B), but it is possible that both directed road <A,B><A,B><A,B> and directed road <B,A><B,A><B,A> exist.
All the test cases are generated randomly.
Output
One line containing a sentence. If it is possible for JOJO to arrive at the destination in time, output "yareyaredawa" (without quote), else output "Whitesnake!" (without quote).
样例输入
2 2
1 2 2 14
1 2 5
2 1 4样例输出
yareyaredawa略有改动：
代码如下
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define N 1100
#define M 100010
#define inf 1000000000
int num1=0,num2=0,n,m,s,t,k,h1[N],h2[N],d[N],flag[N];
struct node1
{
int x,y,z,next;
}mp1[M],mp2[M];
struct node{
int v,c;
node(int vv,int cc) : v(vv),c(cc){}
friend bool operator < (node x,node y)
{
return x.c+d[x.v]>y.c+d[y.v];
}
};
void insert1(int x,int y,int z)
{
mp1[++num1].x=x;
mp1[num1].y=y;
mp1[num1].z=z;
mp1[num1].next=h1[x];
h1[x]=num1;
}
void insert2(int x,int y,int z)
{
mp2[++num2].x=x;
mp2[num2].y=y;
mp2[num2].z=z;
mp2[num2].next=h2[x];
h2[x]=num2;
}
void spfa()
{
queue<int>Q;
Q.push(t);
for(int i=1;i<=n;i++)
d[i]=inf;
d[t]=0;
memset(flag,0,sizeof(flag));
flag[t]=1;
while(!Q.empty())
{
int u=Q.front();Q.pop();
flag[u]=0;
for(int i=h1[u];i;i=mp1[i].next)
{
int v=mp1[i].y;
if(d[v]>d[u]+mp1[i].z)
{
d[v]=d[u]+mp1[i].z;
if(!flag[v])
{
flag[v]=1;
Q.push(v);
}
}
}
}
}
int astar()
{
if(d[s]==inf)
return -1;
priority_queue<node>p;
int cnt=0;
p.push(node(s,0));
while(!p.empty())
{
node u=p.top();p.pop();
if(u.v==t)
{
cnt++;
if(cnt==k)
return u.c;
}
for(int i=h2[u.v];i;i=mp2[i].next)
{
int y=mp2[i].y;
p.push(node(y,u.c+mp2[i].z));
}
}
return -1;
}
int main()
{
int w;
while(scanf("%d%d",&n,&m)>0)
{
scanf("%d%d%d%d",&s,&t,&k,&w);
memset(h1,0,sizeof(h1));
num1=0;
memset(h2,0,sizeof(h2));
num2=0;
for(int i=1;i<=m;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
insert1(y,x,z);
insert2(x,y,z);
}
spfa();
int va = astar();
if(va == -1 || va > w)
printf("Whitesnake!
");
else
printf("yareyaredawa
");
}
return 0;
}