PAT (Advanced Level) Practise 1003 Emergency（SPFA+DFS）

1003. Emergency (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) - the number of cities (and the cities are numbered from 0 to N-1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.
All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input

5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output

2 4

题目链接：1003 Emergency

点似乎不是很多，数据友好，最后一组7MS都能跑完，另外一种感觉麻烦一点但是适应性比较好做法是修改一下dij的写法加上dp的思想，然而不太懂，有空再看……

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=510;
const int M=500010;
struct info
{
	int to;
	int pre;
	int dx;
};
info E[M];
int head[M],ne;
int val[N],d[N],vis[N];
int maxm,ans,n,m,s,t;
void add(int s,int t,int d)
{
	E[ne].to=t;
	E[ne].pre=head[s];
	E[ne].dx=d;
	head[s]=ne++;
}
void init()
{
	CLR(head,-1);
	ne=0;
	CLR(val,0);
	CLR(d,INF);
	ans=0;
	maxm=0;
	CLR(vis,0);
}
void spfa(int s)
{
	int i;
	priority_queue<pii>Q;
	d[s]=0;
	Q.push(pii(-d[s],s));
	while (!Q.empty())
	{
		int now=Q.top().second;
		Q.pop();
		for (i=head[now]; ~i; i=E[i].pre)
		{
			int v=E[i].to;
			if(d[v]>d[now]+E[i].dx)
			{
				d[v]=d[now]+E[i].dx;
				Q.push(pii(-d[v],v));
			}
		}
	}
}
void dfs(int now,int len,int ma)
{
	int i;
	if(now==t)
	{
		if(ma>maxm)
			maxm=ma;
		++ans;
		return ;
	}
	for (i=head[now]; ~i; i=E[i].pre)
	{
		int v=E[i].to;
		if(!vis[v]&&d[now]+E[i].dx==d[v])
		{
			vis[v]=1;
			dfs(v,d[v],ma+val[v]);
			vis[v]=0;
		}
	}
}
int main(void)
{
	int x,y,z,i,j;
	while (~scanf("%d%d%d%d",&n,&m,&s,&t))
	{
		init();
		for (i=0; i<n; ++i)
			scanf("%d",&val[i]);
		for (i=0; i<m; ++i)
		{
			scanf("%d%d%d",&x,&y,&z);
			add(x,y,z);
			add(y,x,z);
		}
		spfa(s);
		dfs(s,d[s],val[s]);
		printf("%d %d\n",ans,maxm);
	}
	return 0;
}

———————10.13更新————————

学会了用Dijsktra来做，特地重新做了一下，发现这题有个坑就是边是双向的，单向边会错几组数据…………

Dij代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=510;
const int M=1e6+7;
struct edge
{
    int to;
    int w;
    int pre;
};
edge E[M];
int head[N],tot;
int d[N],vis[N],way[N],ans[N];
int man[N];
int n,m;
inline void add(int s,int t,int w)
{
    E[tot].to=t;
    E[tot].w=w;
    E[tot].pre=head[s];
    head[s]=tot++;
}
void init()
{
    CLR(head,-1);
    tot=0;
    CLR(d,INF);
    CLR(vis,0);
    CLR(man,0);
    CLR(way,0);
    CLR(ans,0);
}
void Dij(int s)
{
    int i,j,u,v,minm,w;
    d[s]=0;
    vis[s]=1;
    way[s]=1;
    ans[s]=man[s];
    for (i=head[s]; ~i; i=E[i].pre)
    {
        v=E[i].to;
        if(d[v]>E[i].w)
        {
            d[v]=E[i].w;
            way[v]=1;
            ans[v]=ans[s]+man[v];
        }
    }
    for (i=1; i<n; ++i)
    {
        minm=INF;
        u=-1;
        for (j=0; j<n; ++j)
        {
            if(!vis[j]&&d[j]<minm)
            {
                minm=d[j];
                u=j;
            }
        }
        if(u==-1)
            break;
        vis[u]=1;
        for (j=head[u]; ~j; j=E[j].pre)
        {
            v=E[j].to;
            w=E[j].w;
            if(d[v]>d[u]+w)
            {
                d[v]=d[u]+w;
                way[v]=way[u];
                ans[v]=ans[u]+man[v];
            }
            else if(d[v]==d[u]+w)
            {
                way[v]+=way[u];
                ans[v]=max(ans[v],ans[u]+man[v]);
            }
        }
    }
}
int main(void)
{
    int a,b,w,s,t,i;
    while (~scanf("%d%d%d%d",&n,&m,&s,&t))
    {
        init();
        for (i=0; i<n; ++i)
            scanf("%d",&man[i]);
        for (i=0; i<m; ++i)
        {
            scanf("%d%d%d",&a,&b,&w);
            add(a,b,w);
            add(b,a,w);
        }
        Dij(s);
        printf("%d %d\n",way[t],ans[t]);
    }
    return 0;
}