1072. Gas Station_#define n 100001-优快云博客

本文探讨了一种用于解决复杂路径优化问题的算法，包括输入解析、数据结构初始化、算法核心逻辑实现以及输出结果处理。算法通过构建图模型、应用Dijkstra算法寻找最短路径，并考虑了额外约束条件如油量限制、油站可用性等，最终输出最优解。代码实现了从用户输入到结果输出的完整流程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#include<stdio.h>
#include<vector>
#include<string>
#include<cstdlib>
#include<iostream>
using namespace std;

#define N 1012
#define M 100001

struct E{
    int next;
    int dis;
};

vector<E> edge[N];
int dis[11][M];
int sum[11];
bool mark[11][N];
bool chao[11];
bool you[11];

int str2int(char* str,int n){
    int num;
    if(str[0] == 'G'){
        num = atoi(str + 1) + n;
        you[num - n] = true;
    }
    else{
        num = atoi(str);
    }
    return num;
}

int main(){
    int m,n,k,r;
    scanf("%d%d%d%d",&n,&m,&k,&r);
    for(int i = 1;i <= n + m;i ++) edge[i].clear();
    for(int i = 1;i <= 10;i ++) you[i] = false;
    while(k --){
        int x,y,dis;
        char str1[20],str2[20];
        scanf("%s%s%d",str1,str2,&dis);
        x = str2int(str1,n);
        y = str2int(str2,n);

        E tmp;
        tmp.next = y,
        tmp.dis = dis;
        edge[x].push_back(tmp);
        tmp.next = x;
        edge[y].push_back(tmp);

    }

    for(int j = 1;j <= 10;j ++){
        chao[j] = false;
        sum[j] = 0;

    }
    int gasmin[11];

    for(int ii = 1;ii <= m;ii ++){

        int gas = ii;
        int newp = ii + n;
        for(int i = 1;i <= m + n;i ++){
            dis[gas][i] = -1;
            mark[gas][i] = false;
        }

        dis[gas][newp] = 0;
        mark[gas][newp] = true;

        for(int i = 1;i < m + n;i ++){
            for(int j = 0;j < edge[newp].size();j ++){
                int t = edge[newp][j].next;
                int d = edge[newp][j].dis;
                if(mark[gas][t] == true) continue;
                if(dis[gas][t] == -1 || dis[gas][t] > dis[gas][newp] + d){
                    dis[gas][t] = dis[gas][newp] + d;
                }
            }
            int min = 123432434;
            for(int j = 1;j <= m + n;j ++){
                if(mark[gas][j] == true || dis[gas][j] == -1) continue;
                if(min > dis[gas][j]){
                    min = dis[gas][j];
                    newp = j;
                }
            }
            mark[gas][newp] = true;
        }

        sum[gas] = gasmin[gas] = dis[gas][1];
        if(dis[gas][1] > r){
            chao[gas] = true;
            break;
        };
        for(int i = 2;i <= n;i ++){
            if(gasmin[gas] > dis[gas][i]){
                gasmin[gas] = dis[gas][i];
            }
            if(dis[gas][i] > r){
                chao[gas] = true;
                break;
            }
            sum[gas] += dis[gas][i];
        }
    }

    int ans = 1;
    for(int i = 2;i <= m;i ++){
        if(chao[i] == true || you[i] == false) continue;
        if(gasmin[ans] < gasmin[i] || gasmin[ans] == gasmin[i] && sum[ans] > sum[i]){
            ans = i;
            continue;
        }
    }
    if(chao[ans] == true || you[ans] == false ||gasmin[ans] == -1)
        cout<<"No Solution"<<endl;
    else{
        cout<<"G"<<ans<<endl;
        printf("%.1f %.1f\n",gasmin[ans]*1.0,sum[ans]*1.0/n);
    }
    return 0;
}