This question is really a good question, it basically covers all the test sites in Dijkstral.
I calmed down and wrote for an hour, and found that test points 5 and 7 could not pass. I analyzed it for a while to no avail. I consulted other people’s blogs and found that these two points are pits, because the bicycles behind the road cannot fill the lack of the bicycles in front. The extra bicycles in the front can fill in the missing ones in the back.
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int INF = 1e9;
const int MAXV = 510;
int d[MAXV];
int num[MAXV];
int G[MAXV][MAXV];
int half;
int tb = INF;
int mn = INF;
int out = 0;
int in = 0;
int c, n, p, m;
bool vis[MAXV] = {
false};
vector<int> pre[MAXV], tempPath, path;
void Dijkstral(int s){
fill(d, d+MAXV, INF);
d[s] = 0;
for(int i=0; i<=n; i++){
int u = -1, MIN = INF;
for(int j=0; j<=n; j++){
if(vis[j]==false && d[j]<MIN){
u = j;
MIN = d[j];
}
}
if(u == -1) return;
vis[u] = true;
for(int v=0; v<=n; v++){
if(vis[v]==false && G[u][v]!=INF){
if(d[v] > d[u] + G[u][v]){
d[v] = d[u] + G[u][v];
pre[v].clear();
pre[v].push_back(u);
}else if(d[v] == d[u] + G[u][v]){
pre[v].push_back(u);
}
}
}
}
}
void DFS(int e){
if(e == 0){
tempPath.push_back(e);
int need = 0;
int back = 0;
for(int i=0; i<tempPath.size()-1; i++){
int index = tempPath[i];
if(num[index] > half){
back += (num[index]-half);
if(need != 0){
if(back <= need){
need -= back;
back = 0;
}else{
back -= need;
need = 0;
}
}
}else if(num[index] < half){
need += (half-num[index]);
}
}
if(need < mn){
tb = back;
path = tempPath;
in = tb;
mn = need;
out = need;
}else if(need == mn){
if(back < tb){
tb = back;
path = tempPath;
in = tb;
mn = need;
out = need;
}
}
tempPath.pop_back();
}
tempPath.push_back(e);
for(int i=0; i<pre[e].size(); i++){
DFS(pre[e][i]);
}
tempPath.pop_back();
}
int main(){
fill(G[0], G[0]+MAXV*MAXV, INF);
int u, v, w;
scanf("%d %d %d %d", &c, &n, &p, &m);
for(int i=1; i<=n; i++){
scanf("%d", &num[i]);
}
half = c/2;
for(int i=0; i<m; i++){
scanf("%d %d %d", &u, &v, &w);
G[u][v] = w;
G[v][u] = w;
}
Dijkstral(0);
DFS(p);
printf("%d ", out);
for(int i=path.size()-1; i>=0; i--){
printf("%d", path[i]);
if(i != 0) printf("->");
}
printf(" %d", in);
return 0;
}