Dijkstra專題近三年考察情況：19年9月第四題（題解在此：https://blog.csdn.net/allisonshing/article/details/100643092）
筆記：
liuchuo滿分代碼展示
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;
//下面兩個函數將記錄前驅節點的數組轉換成逆序的序列放到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);//地圖中地點的個數和路路徑的條數
int a, b, flag, len, t;
for(int i = 0; i < m; i++) {
//道路路結點編號V1 道路路結點編號V2 是否單行線 道路長度 所需時間
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]) {//求最短路徑
dis[v] = e[u][v] + dis[u];
dispre[v] = u;//v的前驅是u
weight[v] = weight[u] + w[u][v];
} else if(e[u][v] + dis[u] == dis[v] && weight[v] > weight[u] + w[u][v]) {//如果相同求時間最短的那條
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++) {//求最快路徑（如果相同求結點數最⼩的那條）
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]) {//如果相同求結點數最⼩的
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
這道題的背景開始讓我挺困惑的，不過也能明白：
好比是我要給我的車加油，我會選最偏僻郊區（離居民區最近的人最遠）但是還在我能到的範圍內的一座加油站。
而且如果有多個相同距離的加油站，就選離我的鄰居們平均最近的一個。
“離居民區最近的人最遠”有兩個最字，就是比較兩次最值，第一次是加油站離居民區最近的距離，第二次是在第一次基礎上，找最遠距離的加油站。
第三個最字在“離我的鄰居們平均最近”，這個和上面兩個最字裏一起統計。
#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++) {//計算每個加油站到居民區每戶人家的距離
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++) {//計算每戶居民或者加油站出發到別的點的最短路徑，不要區別是居民還是加油站
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++) {//找出離該處加油站最近的居民
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) {//找出離最近的居民最遠的加油站
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;
}