(此处有前向星删边操作)先提供每删一条免费机票的边就跑一遍迪杰斯特拉,不过由于每次都跑一遍,复杂度较大需要堆优化。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cstdlib>
#include<cmath>
#include<stack>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
const double pi = acos(-1);
const int maxn = 1200;
const ll mod = 1e9 + 7;
int n, s, t, k, q, cnt;
int head[maxn], vis[maxn], dis[maxn];
struct node{
int to;
int quan;
int next;
}edge[maxn * maxn << 4];
struct nod{
int u;
int d;
bool operator<(const nod& xgd) const{
return d > xgd.d;
}
};
priority_queue<nod> que;
void add(int u, int v, int w){
edge[++cnt].to = v;
edge[cnt].quan = w;
edge[cnt].next = head[u];
head[u] = cnt;
}
int dijiskra(int s, int t){
memset(dis, INF, sizeof(dis));
memset(vis, 0, sizeof(vis));
que.push(nod{s, 0});
dis[s] = 0;
while(!que.empty()){
nod cur = que.top();
que.pop();
int k = cur.u;
if(vis[k]) continue;
vis[k] = 1;
for(int i = head[k] ; i != -1 ; i = edge[i].next){
int tt = edge[i].to;
if(!vis[tt] && dis[tt] > dis[k] + edge[i].quan){
dis[tt] = dis[k] + edge[i].quan;
que.push(nod{tt, dis[tt]});
}
}
}
return dis[t];
}
int main()
{
memset(head, -1, sizeof(head));
int u, v, w;
scanf("%d %d %d", &n, &s, &t);
scanf("%d", &k);
for(int i = 1 ; i <= k ; ++ i){
scanf("%d %d %d", &u, &v, &w);
add(u, v, w); add(v, u, w);
}
int ans = dijiskra(s, t);
int anss = INF;
scanf("%d", &q);
for(int i = 1 ; i <= q ; ++ i){
scanf("%d %d", &u, &v);
add(u, v, 0); add(v, u, 0);
anss = min(anss, dijiskra(s, t));
head[v] = edge[cnt].next; cnt --;
head[u] = edge[cnt].next; cnt --;
}
if(ans > anss) cout << "Yes" << endl;
else cout << "No" << endl;
cout << anss;
return 0;
}
直接正向反向跑一遍迪杰斯特拉,dis代表正向跑, diss代表反向跑,最后最短的距离一定是min(dis[u]+diss[v],dis[v]+diss[u], dis[t]).
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cstdlib>
#include<cmath>
#include<stack>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
const double pi = acos(-1);
const int maxn = 1200;
const ll mod = 1e9 + 7;
int n, s, t, k, q, cnt;
int head[maxn], vis[maxn], dis[maxn];
struct node{
int to;
int quan;
int next;
}edge[maxn * maxn << 4];
struct nod{
int u;
int d;
bool operator<(const nod& xgd) const{
return d > xgd.d;
}
};
priority_queue<nod> que;
void add(int u, int v, int w){
edge[cnt].to = v;
edge[cnt].quan = w;
edge[cnt].next = head[u];
head[u] = cnt ++;
}
int dijiskra(int s, int t){
memset(dis, INF, sizeof(dis));
memset(vis, 0, sizeof(vis));
que.push(nod{s, 0});
dis[s] = 0;
while(!que.empty()){
nod cur = que.top();
que.pop();
int k = cur.u;
if(vis[k]) continue;
vis[k] = 1;
for(int i = head[k] ; i != -1 ; i = edge[i].next){
int tt = edge[i].to;
if(!vis[tt] && dis[tt] > dis[k] + edge[i].quan){
dis[tt] = dis[k] + edge[i].quan;
que.push(nod{tt, dis[tt]});
}
}
}
return dis[t];
}
int main()
{
memset(head, -1, sizeof(head));
int u, v, w;
scanf("%d %d %d", &n, &s, &t);
scanf("%d", &k);
for(int i = 1 ; i <= k ; ++ i){
scanf("%d %d %d", &u, &v, &w);
add(u, v, w); add(v, u, w);
}
int ans = dijiskra(s, t);
int anss = INF;
scanf("%d", &q);
for(int i = 1 ; i <= q ; ++ i){
scanf("%d %d", &u, &v);
add(u, v, 0); add(v, u, 0);
anss = min(anss, dijiskra(s, t));
head[v] = edge[--cnt].next;
head[u] = edge[--cnt].next;
}
if(ans > anss) cout << "Yes" << endl;
else cout << "No" << endl;
cout << anss;
return 0;
}