Dijkstra专题近三年考察情况：19年9月第四题（题解在此：https://blog.csdn.net/allisonshing/article/details/100643092）

笔记：

liuchuo满分代码展示

1111 Online Map (30分)

https://pintia.cn/problem-sets/994805342720868352/problems/994805358663417856

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 999999999;
int dis[510], Time[510], e[510][510], w[510][510], dispre[510],Timepre[510], weight[510],NodeNum[510];
bool visit[510];
vector<int> dispath, Timepath, temppath;
int st, fin, minnode = inf;
//下面两个函数将记录前驱节点的数组转换成逆序的序列放到vector
void dfsdispath(int v) {
    dispath.push_back(v);
    if(v == st) return ;
    dfsdispath(dispre[v]);
}
void dfsTimepath(int v) {
    Timepath.push_back(v);
    if(v == st) return ;
    dfsTimepath(Timepre[v]);
}
int main() {
    fill(dis, dis + 510, inf);
    fill(Time, Time + 510, inf);
    fill(weight, weight + 510, inf);
    fill(e[0], e[0] + 510 * 510, inf);
    fill(w[0], w[0] + 510 * 510, inf);
    int n, m;
    scanf("%d %d", &n, &m);//地图中地点的个数和路路径的条数
    int a, b, flag, len, t;
    for(int i = 0; i < m; i++) {
        //道路路结点编号V1 道路路结点编号V2 是否单行线 道路长度 所需时间
        scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);
        e[a][b] = len;
        w[a][b] = t;
        if(flag != 1) {
            e[b][a] = len;
            w[b][a] = t;
        }
    }
    scanf("%d %d", &st, &fin);
    dis[st] = 0;
    for(int i = 0; i < n; i++)
        dispre[i] = i;
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && dis[j] < minn) {
                u = j;
                minn = dis[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && e[u][v] != inf) {
                if(e[u][v] + dis[u] < dis[v]) {//求最短路径
                    dis[v] = e[u][v] + dis[u];
                    dispre[v] = u;//v的前驱是u
                    weight[v] = weight[u] + w[u][v];
                } else if(e[u][v] + dis[u] == dis[v] && weight[v] > weight[u] + w[u][v]) {//如果相同求时间最短的那条
                    weight[v] = weight[u] + w[u][v];
                    dispre[v] = u;
                }
            }
        }
    }
    dfsdispath(fin);
    Time[st] = 0;
    fill(visit, visit + 510, false);
    for(int i = 0; i < n; i++) {//求最快路径（如果相同求结点数最⼩的那条）
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && minn > Time[j]) {
                u = j;
                minn = Time[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && w[u][v] != inf) {
                if(w[u][v] + Time[u] < Time[v]) {
                    Time[v] = w[u][v] + Time[u];
                    Timepre[v]=(u);
                    NodeNum[v]=NodeNum[u]+1;
                } else if(w[u][v] + Time[u] == Time[v]&&NodeNum[u]+1<NodeNum[v]) {//如果相同求结点数最⼩的
                    Timepre[v]=(u);
                    NodeNum[v]=NodeNum[u]+1;
                }
            }
        }
    }
    dfsTimepath(fin);
    printf("Distance = %d", dis[fin]);
    if(dispath == Timepath) {
        printf("; Time = %d: ", Time[fin]);
    } else {
        printf(": ");
        for(int i = dispath.size() - 1; i >= 0; i--) {
            printf("%d", dispath[i]);
            if(i != 0) printf(" -> ");
        }
        printf("\nTime = %d: ", Time[fin]);
    }
    for(int i = Timepath.size() - 1; i >= 0; i--) {
        printf("%d", Timepath[i]);
        if(i != 0) printf(" -> ");
    }
    return 0;
}

1072. Gas Station (30)

https://pintia.cn/problem-sets/994805342720868352/problems/994805396953219072

这道题的背景开始让我挺困惑的，不过也能明白：

好比是我要给我的车加油，我会选最偏僻郊区（离居民区最近的人最远）但是还在我能到的范围内的一座加油站。

而且如果有多个相同距离的加油站，就选离我的邻居们平均最近的一个。

“离居民区最近的人最远”有两个最字，就是比较两次最值，第一次是加油站离居民区最近的距离，第二次是在第一次基础上，找最远距离的加油站。

第三个最字在“离我的邻居们平均最近”，这个和上面两个最字里一起统计。

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
const int inf = 999999999;
int n, m, k, ds, station;
int e[1020][1020], dis[1020]; 
bool visit[1020];
int main() {
    fill(e[0], e[0] + 1020 * 1020, inf);
    fill(dis, dis + 1020, inf);
    scanf("%d%d%d%d", &n, &m, &k, &ds);
    for(int i = 0; i < k; i++) {
        int tempdis;
        string s, t;
        cin >> s >> t >> tempdis;
        int a, b;
        if(s[0] == 'G') {
            s = s.substr(1);
            a = n + stoi(s);
        } else {
            a = stoi(s);
        }
        if(t[0] == 'G') {
            t = t.substr(1);
            b = n + stoi(t);
        } else {
            b = stoi(t);
        }
        e[a][b] = e[b][a] = tempdis;
    }
    int ansid = -1;
    double ansdis = -1, ansaver = inf;
    for(int index = n + 1; index <= n + m; index++) {//计算每个加油站到居民区每户人家的距离
        double mindis = inf, aver = 0;
        fill(dis, dis + 1020, inf);
        fill(visit, visit + 1020, false);
        dis[index] = 0;
        for(int i = 0; i < n + m; i++) {//计算每户居民或者加油站出发到别的点的最短路径，不要区别是居民还是加油站
            int u = -1, minn = inf;
            for(int j = 1; j <= n + m; j++) {
                if(visit[j] == false && dis[j] < minn) {
                    u = j;
                    minn = dis[j];
                }
            }
            if(u == -1) break;
            visit[u] = true;
            for(int v = 1; v <= n + m; v++) {
                if(visit[v] == false && dis[v] > dis[u] + e[u][v])
                    dis[v] = dis[u] + e[u][v];
            }
        }
        for(int i = 1; i <= n; i++) {//找出离该处加油站最近的居民
            if(dis[i] > ds) {
                mindis = -1;
                break;
            }
            if(dis[i] < mindis) mindis = dis[i];
            aver += 1.0 * dis[i];
        }
        if(mindis == -1) continue;
        aver = aver / n;
        if(mindis > ansdis) {//找出离最近的居民最远的加油站
            ansid = index;
            ansdis = mindis;
            ansaver = aver;
        } else if(mindis == ansdis && aver < ansaver) {
            ansid = index;
            ansaver = aver;
        }
    }
    if(ansid == -1)
        printf("No Solution");
    else
        printf("G%d\n%.1f %.1f", ansid - n, ansdis, ansaver);
    return 0;
}

1003 Emergency (25分)

https://pintia.cn/problem-sets/994805342720868352/problems/994805523835109376

#include <iostream>
#include <algorithm>
using namespace std;
int n, m, c1, c2;
int e[510][510], weight[510], dis[510], num[510], w[510];
bool visit[510];
const int inf = 99999999;
int main() {
    scanf("%d%d%d%d", &n, &m, &c1, &c2);
    for(int i = 0; i < n; i++)
        scanf("%d", &weight[i]);
    fill(e[0], e[0] + 510 * 510, inf);
    fill(dis, dis + 510, inf);
    int a, b, c;
    for(int i = 0; i < m; i++) {
        scanf("%d%d%d", &a, &b, &c);
        e[a][b] = e[b][a] = c;
    }
    dis[c1] = 0;
    w[c1] = weight[c1];
    num[c1] = 1;
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && dis[j] < minn) {
                u = j;
                minn = dis[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && e[u][v] != inf) {
                if(dis[u] + e[u][v] < dis[v]) {
                    dis[v] = dis[u] + e[u][v];
                    num[v] = num[u];
                    w[v] = w[u] + weight[v];
                } else if(dis[u] + e[u][v] == dis[v]) {
                    num[v] = num[v] + num[u];
                    if(w[u] + weight[v] > w[v])
                        w[v] = w[u] + weight[v];
                }
            }
        }
    }
    printf("%d %d", num[c2], w[c2]);
    return 0;
}