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POJ 3159 Candies(差分约束+dijkstra)

2016-11-20 09:15 260 查看
根据B-A<=C可以从A连一条权值为C的边,然后从差分式子中可以看出这是求最短路,因为没有负权边,所以用dijkstra更快,但是vector建图会超时,所以用前向星建图。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int maxn=30000+5;
const int maxc=150000+5;
const int inf=100000000;
struct HeapNode{
int d,u;
bool operator < (const HeapNode &rhs) const {
return d>rhs.d;
}
};
struct edge{
int from,to,dist;
};
struct Dijkstra{
int n,m;
edge edges[maxc];
int head[maxc];
int next[maxc];
bool done[maxn];
int d[maxn];
int p[maxn];

void init(int n){
this->n=n;
memset(head,-1,sizeof(head));
memset(next,-1,sizeof(next));
}

void addedge(int from,int to,int dist){
edges[m]=(edge){from,to,dist};
next[m]=head[from];
head[from]=m++;
}

void dijkstra(int s){
priority_queue<HeapNode> pq;
for(int i=0;i<n;i++) d[i]=inf;
d[s]=0;
pq.push((HeapNode){0,s});
memset(done,0,sizeof(done));
while(!pq.empty()){
HeapNode x=pq.top();pq.pop();
int u=x.u;
if(done[u]) continue;
done[u]=true;
for(int i=head[u];i!=-1;i=next[i]){
edge &e=edges[i];
if(d[e.to]>d[u]+e.dist){
d[e.to]=d[u]+e.dist;
p[e.to]=i;
pq.push((HeapNode){d[e.to],e.to});
}
}
}
}
};

Dijkstra dij;
int n,m;
int main()
{
while(~scanf("%d%d",&n,&m))
{
dij.init(n);
int u,v,dist;
for(int i=0;i<m;i++){
scanf("%d%d%d",&u,&v,&dist);
u--;v--;
dij.addedge(u,v,dist);
}
dij.dijkstra(0);
printf("%d\n",dij.d[n-1]);
}
return 0;
}
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