【PTAL2-001】紧急救援（Dijkstra+最短路径的条数+最短路径中点权之和的最大值）

题目地址：https://pintia.cn/problem-sets/994805046380707840/problems/994805073643683840

题目：

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解题思路：

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dijkstra

num[i]表示到结点i的最短路径条数

sum[i]表示到结点i的（最短路径中）能召集的最多救援队数量

当在更新最短路的时候遇到：（以u为中介，s->u+u->v==s->v）

dis[v] == dis[u] + edges[u][v]

最短路径的条数为num[u]+num[v]，如果peo[v]+sum[u]>sum[v]则更新sum[v]

 

ac代码：

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dijkstra中循环n次（不直接初始化dis[i]）推荐这样写，代码量少，更简单

#include <iostream>
#include <algorithm>
#include <string.h>
#include <ctype.h>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <sstream>
#define  maxn 505
typedef long long ll;
const ll inf=99999999;
using namespace std;
ll dis[maxn],edges[maxn][maxn],pre[maxn],peo[maxn],vis[maxn]={0},sum[maxn]={0},num[maxn]={0};
ll n,m,s,d,a,b,c;
vector<ll> ans;
void solve()
{
    edges[s][s]=0;dis[s]=0;
    sum[s]=peo[s],num[s]=1;
    for(int j=0;j<n;j++)
    {
        ll min = inf - 1, u = -1;
        for (ll i = 0; i < n; i++)
        {
            if (dis[i] < min && !vis[i])
            {
                min = dis[i];
                u = i;
            }
        }
        if(u==-1) return ;//不连通
        vis[u] = 1;
        for (ll v = 0; v < n; v++)
        {
            if(!vis[v]&&edges[u][v]!=inf)
            {

                if (dis[v] > dis[u] + edges[u][v])
                {
                    dis[v] = dis[u] + edges[u][v];
                    pre[v] = u;
                    sum[v] = peo[v] + sum[u];//更新获取的救援队的数目
                    num[v] = num[u];//更新到达i最短路的条数
                } else if (dis[v] == dis[u] + edges[u][v])
                {
                    num[v] += num[u];//更新条数
                    if (peo[v] + sum[u] > sum[v])//更新后的sum[i]>原sum[i]
                    {
                        sum[v] = peo[v] + sum[u];
                        pre[v] = u;
                    }
                }
            }
        }
    }
}
int main()
{
    //freopen("/Users/zhangkanqi/Desktop/11.txt","r",stdin);
    scanf("%lld %lld %lld %lld",&n,&m,&s,&d);
    fill(edges[0],edges[0]+maxn*maxn,inf);
    fill(dis,dis+maxn,inf);
    for(ll i=0;i<n;i++)
        scanf("%lld",&peo[i]);
    for(ll i=0;i<m;i++)
    {
        scanf("%lld %lld %lld",&a,&b,&c);
        edges[a][b]=edges[b][a]=c;
    }
    solve();
    ll x=d;
    while(x!=s)
    {
        ans.push_back(x);
        x=pre[x];
    }
    ans.push_back(s);
    printf("%lld %lld\n",num[d],sum[d]);
    for(int i=ans.size()-1;i>=0;i--)
    {
        if(i!=ans.size()-1)
            printf(" ");
        printf("%lld",ans[i]);
    }
    return 0;
}

dijkstra中循环n-1次（dis[i]的初始化放在dijkstra的外面写）

#include <iostream>
#include <algorithm>
#include <string.h>
#include <ctype.h>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <sstream>
#define  maxn 505
typedef long long ll;
const ll inf=99999999;
using namespace std;
ll dis[maxn],edges[maxn][maxn],pre[maxn],peo[maxn],vis[maxn]={0},sum[maxn]={0},num[maxn]={0};
ll n,m,s,d,a,b,c;
vector<ll> ans;
void solve()
{
    for(int j=1;j<n;j++)
    {
        ll min = inf - 1, u = -1;
        for (ll i = 0; i < n; i++)
        {
            if (dis[i] < min && !vis[i])
            {
                min = dis[i];
                u = i;
            }
        }
        if(u==-1) return ;//不连通
        vis[u] = 1;
        for (ll v = 0; v < n; v++)
        {
            if(!vis[v]&&edges[u][v]!=inf)
            {

                if (dis[v] > dis[u] + edges[u][v])
                {
                    dis[v] = dis[u] + edges[u][v];
                    pre[v] = u;
                    sum[v] = peo[v] + sum[u];//更新获取的救援队的数目
                    num[v] = num[u];//更新到达i最短路的条数
                } else if (dis[v] == dis[u] + edges[u][v])
                {
                    num[v] += num[u];//更新条数
                    if (peo[v] + sum[u] > sum[v])//更新后的sum[i]>原sum[i]
                    {
                        sum[v] = peo[v] + sum[u];
                        pre[v] = u;
                    }
                }
            }
        }
    }
}
int main()
{
    //freopen("/Users/zhangkanqi/Desktop/11.txt","r",stdin);
    scanf("%lld %lld %lld %lld",&n,&m,&s,&d);
    fill(edges[0],edges[0]+maxn*maxn,inf);
    fill(dis,dis+maxn,inf);
    for(ll i=0;i<n;i++)
        scanf("%lld",&peo[i]);
    for(ll i=0;i<m;i++)
    {
        scanf("%lld %lld %lld",&a,&b,&c);
        edges[a][b]=edges[b][a]=c;
    }
    edges[s][s]=0;dis[s]=0;
    sum[s]=peo[s],num[s]=1;
    vis[s]=1;
    for(int i=0;i<n;i++)
        if(edges[s][i]!=inf&&i!=s)
        {
            dis[i]=edges[s][i];
            pre[i]=s;
            num[i]=num[s];
            sum[i]=sum[s]+peo[i];
        }
    solve();
    ll x=d;
    while(x!=s)
    {
        ans.push_back(x);
        x=pre[x];
    }
    ans.push_back(s);
    printf("%lld %lld\n",num[d],sum[d]);
    for(int i=ans.size()-1;i>=0;i--)
    {
        if(i!=ans.size()-1)
            printf(" ");
        printf("%lld",ans[i]);
    }
    return 0;
}