单点时限: 2.0 sec
内存限制: 256 MB
Kid has got the diamond ring after he got the password. After that he tried to escape as quickly as possible. Now give you the map, point 0 stands for Kid, point 1 stands for the exit.
输入格式
There are n (2<=n<=100) points (from 0 to n-1), r roads, given in a line. Different distances between points are given, one line for each road with three integers s, e, l(0<=s, e<=n-1, 0<l<=1000), stands for the road between the point s and e, length l.
输出格式
Please find out the shortest way for Kid to escape. Just output the length of the way you’ve found.
样例
input
5 6
0 2 2
0 3 2
2 3 3
0 4 12
3 1 34
4 1 24
output
36
思路：Dijkstra单元最短路径
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
struct g {
int x;
int cost;
friend bool operator < (g a,g b) {
return a.cost>b.cost;
}
};
const int inf=100000;
bool visit[100]= {false};//是否访问
bool visit1[100]= {false};//判断是否入队
int d[100];
int n;
vector<g>v[100];
void Dijkstra_1(int s) {
fill(d,d+100,inf);
d[s]=0;
for(int i = 0; i < n; i++) {
int u=-1,Min=inf;
for(int j = 0; j < n; j++) {
if(visit[j]==false&&Min>d[j]) {
u=j;
Min=d[j];
}
}
if(u==-1)
return ;
visit[u]=true;
for(int i = 0; i < v[u].size(); i++) {
int w=v[u][i].x;
int cost=v[u][i].cost;
if(visit[w]==false&&d[u]+cost<d[w]) {
d[w]=d[u]+cost;
}
}
}
}
void Dijkstra_2(int s) {//优先队列实现
fill(d,d+100,inf);
d[s]=0;
priority_queue<g>q;
g G;
G.x=s;
G.cost=0;
q.push(G);
visit1[s]=true;//入队
while(!q.empty()) {
g f=q.top();
q.pop();
visit1[f.x]=false;//不在队列
for(int i = 0; i < v[f.x].size(); i++) {
int u=v[f.x][i].x;
int cost=v[f.x][i].cost;
if(visit[u]==false&&d[f.x]+cost<d[u]) {
d[u]=d[f.x]+cost;
if(visit1[u]==false) {//为入队
g m;
m.x=u;
m.cost=v[f.x][i].cost;
q.push(m);
visit1[u]=true;
}
}
}
}
}
int main() {
int m;
cin>>n>>m;
for(int i = 0; i < m; i++) {
int a,b,cost;
cin>>a>>b>>cost;
g G;
G.x=b;
G.cost=cost;
v[a].push_back(G);
G.x=a;
v[b].push_back(G);
}
Dijkstra_2(0);
cout<<d[1];
// for(int i = 0; i < n; i++)
// cout<<d[i]<<" ";
return 0;
}