USACO 2010 Feb Chocolate Giving 领取巧克力

这道题其实就是堆优化的Dijkstra。不用堆优化的话是会超时的。

#include<cstdio>
#include<cstdlib>

#define PROBNAME "cgiving"

const int MAXN=50000;
const int MAXM=100000;
const int MAXB=25000;
const int INF=2000*MAXN*2+1;

int n,m,b;
int k,first[MAXN+2],next[MAXM*2+10],head[MAXM*2+10],tail[MAXM*2+10],c[MAXM*2+10];
int q[MAXN+2],rd[MAXN+2],size;
int d[MAXN+2];

void open_file(void)
{
	freopen(PROBNAME".in","r",stdin);
	freopen(PROBNAME".out","w",stdout);
}

int getint(void)
{
	int a;
	scanf("%d",&a);
	return a;
}

void addedge(int u,int v,int w)
{
	k++;
	next[k]=first[u];
	first[u]=k;
	head[k]=u;
	tail[k]=v;
	c[k]=w;
}

void init(void)
{
	n=getint();
	m=getint();
	b=getint();
	for(int i=1;i<=m;i++)
	{
		int r=getint(),s=getint(),l=getint();
		addedge(r,s,l);
		addedge(s,r,l);
	}
}

void swap(int i, int j) {
  int u = q[i];
  int v = q[j];
  rd[u] = j;  rd[v] = i;
  q[i] = v;  q[j] = u;
}

void up(int i)
{
	int j=i/2;
	if(j>0&&d[q[i]]<d[q[j]])
	{
		swap(i,j);
		up(j);
	}
}

void down(int i)
{
	int j=i*2;
	if(j+1<=size&&d[q[j]]>d[q[j+1]])
		j++;
	if(j<=size&&d[q[i]]>d[q[j]])
	{
		swap(i,j);
		down(j);
	}
}

void push(int x)
{
	q[++size]=x;
	rd[x]=size;
	up(size);
}

void pop(void)
{
	swap(1,size--);
	down(1);
}

void dijkstra(void)
{
	for(int i=2;i<=n;i++)
	{
		d[i]=INF;
	}
	for(int i=1;i<=n;i++)
	{
		push(i);
	}
	while(size)
	{
		int u=q[1];
		pop();
		for(int e=first[u];e;e=next[e])
		{
			int v=tail[e];
			if(d[v]>d[u]+c[e])
			{
				d[v]=d[u]+c[e];
				up(rd[v]);
			}
		}
	}
}

void solve(void)
{
	dijkstra();
}

void output(void)
{
	for(int i=1;i<=b;i++)
	{
		int p=getint(),q=getint();
		printf("%d\n",d[p]+d[q]);
	}
}

void close_file(void)
{
	fclose(stdin);
	fclose(stdout);
}

int main()
{
	open_file();
	init();
	solve();
	output();
	close_file();
	return 0;
}