https://www.patest.cn/contests/pat-a-practise/1087
在求最短路径种数的时候,是这样写的:
#include
using namespace std;
int n,m,ed,a;
string st,z,y;
map r;
struct node{
string a;
int b;
}x[205];
int w[205][205];
int path[205];
int num[205];
int happy[205];
int cost[205];
int main(){
cin>>n>>m>>st;
for(int i=0;i>x[i].a>>x[i].b;
if(x[i].a=="ROM")
ed=i;
r[x[i].a]=i;
}
while(m--){
cin>>y>>z>>a;
w[r[y]][r[z]]=a;
w[r[z]][r[y]]=a;
}
for(int i=0;i q;
q.push(0);
cost[0]=0;
int costs=100000000,happys=-1,nums=100000000,sum=-1,p;
while(!q.empty()){
int fr=q.top();
q.pop();
if(fr==ed){
if(cost[fr]happys)||
(cost[fr]==costs&&happy[fr]==happys&&num[fr]happy[i]||
(happy[fr]+x[i].b==happy[i]&&num[fr]+1 st;
st.push(ed);
while(ed!=0){
ed=path[ed];
st.push(ed);
}
while(!st.empty()){
cout<";
}
}
然后就因为这个【
最短路径种数】和【内存超限】报错了。
修改之后是这样写的:
#include
using namespace std;
int n,m,ed,a;
string st,z,y;
map r;
struct node{
string a;
int b;
}x[205];
int w[205][205];
int path[205];
int num[205];
int happy[205];
int cost[205];
int over[205];
int main(){
cin>>n>>m>>st;
for(int i=0;i>x[i].a>>x[i].b;
if(x[i].a=="ROM")
ed=i;
r[x[i].a]=i;
}
while(m--){
cin>>y>>z>>a;
w[r[y]][r[z]]=a;
w[r[z]][r[y]]=a;
}
for(int i=0;i q;
q.push(0);
cost[0]=0;
over[0]=1;
int costs=100000000,happys=-1,nums=100000000,sum=-1,p;
while(!q.empty()){
int fr=q.top();
q.pop();
if(fr==ed){
if(cost[fr]happys)||
(cost[fr]==costs&&happy[fr]==happys&&num[fr]happy[i]||
(happy[fr]+x[i].b==happy[i]&&num[fr]+1 st;
st.push(ed);
while(ed!=0){
ed=path[ed];
st.push(ed);
}
while(!st.empty()){
cout<";
}
}
一开始以为【内存超限】是因为我没有把费用相同的路径合并计算(即第一个代码),但其实不是这个原因,而是我的Dijkstra是优先队列nlogn版的……改成了n2的版本就过了。
#include
using namespace std;
int n,m,ed,a;
string st,z,y;
map r;
struct node{
string a;
int b;
}x[205];
int w[205][205];
int path[205];
int num[205];
int happy[205];
int cost[205];
int over[205];
int main(){
cin>>n>>m>>st;
for(int i=0;i>x[i].a>>x[i].b;
if(x[i].a=="ROM")
ed=i;
r[x[i].a]=i;
}
while(m--){
cin>>y>>z>>a;
w[r[y]][r[z]]=a;
w[r[z]][r[y]]=a;
}
for(int i=0;ihappys)||
(cost[fr]==costs&&happy[fr]==happys&&num[fr]happy[i])||
(cost[fr]+w[fr][i]==cost[i]&&happy[fr]+x[i].b==happy[i]&&num[fr]+1 st;
st.push(ed);
while(ed!=0){
ed=path[ed];
st.push(ed);
}
while(!st.empty()){
cout<";
}
}
#include
using namespace std;
int n,m,ed,a;
string st,z,y;
map r;
struct node{
string a;
int b;
}x[205];
int w[205][205];
int path[205];
int num[205];
int happy[205];
int cost[205];
int over[205];
struct cmp{
bool operator()(const int &t1,const int &t2){
return cost[t1]>cost[t2]; //从小到小=大,与数组规则相反
}
};
int main(){
cin>>n>>m>>st;
for(int i=0;i>x[i].a>>x[i].b;
if(x[i].a=="ROM")
ed=i;
r[x[i].a]=i;
}
while(m--){
cin>>y>>z>>a;
w[r[y]][r[z]]=a;
w[r[z]][r[y]]=a;
}
for(int i=0;i,cmp> q;
q.push(0);
cost[0]=0;
over[0]=1;
int vis[205]={0}; //这步千万别忘啊!!
int costs=100000000,happys=-1,nums=100000000,p;
while(!q.empty()){
int fr=q.top();
q.pop();
if(vis[fr]==1) continue; //这步千万别忘啊!!
vis[fr]=1; //这步千万别忘啊!!
if(fr==ed){
if(cost[fr]happys)||
(cost[fr]==costs&&happy[fr]==happys&&num[fr]happy[i])||
(cost[fr]+w[fr][i]==cost[i]&&happy[fr]+x[i].b==happy[i]&&num[fr]+1 st;
st.push(ed);
while(ed!=0){
ed=path[ed];
st.push(ed);
}
while(!st.empty()){
cout<";
}
}
总结一下:考PAT还是别写什么复杂的优先队列还是前向星了,节省时间增加正确率才是王道!
另外,PAT的最短路做来做去感觉好像都差不多。比如:
https://www.patest.cn/contests/pat-a-practise/1111
#include
using namespace std;
struct node{
int a,b;
}x[505][505];
int dis[505];
int spd[505];
int vis[505];
int path[505];
int path2[505];
int n,m,a,b,c,d,e,st,ed;
int main(){
cin>>n>>m;
for(int i=0;i>a>>b>>c>>d>>e;
if(c==1)
x[a][b]={d,e};
else{
x[a][b]={d,e};
x[b][a]={d,e};
}
}
cin>>st>>ed;
int tmp=ed;
for(int i=0;i stk;
if(u==0){
cout<<"Distance = "< ";
else
cout< ";
else
cout< ";
else
cout<
再比如:
https://www.patest.cn/contests/gplt/L2-001
#include
#define ll long long
using namespace std; //
struct EDGE{
int u,v,w,next;
}edge[500*500*2+5];
int head[500*500+5],pp;
void init(){
pp=0;
memset(head,0,sizeof(head));
}
void add(int u,int v,int w){
edge[++pp]=(EDGE){u,v,w,head[u]};
head[u]=pp;
}
int n,m,a,b,c,d,e;
int x[505];
int vis[505];
int dis[505];
int cost[505];
int path[505];
int times[505];
struct cmp{
bool operator()(const int &t1,const int &t2){
if(dis[t1]!=dis[t2])
return dis[t1]>dis[t2];
return cost[t1]>n>>m>>a>>b;
init();
for(int i=0;i>x[i];
while(m--){
cin>>c>>d>>e;
add(c,d,e);
add(d,c,e);
}
for(int i=0;i,cmp> q;
q.push(a);
dis[a]=0;
cost[a]=x[a];
times[a]=1;
while(!q.empty()){
int u=q.top();
q.pop();
if(vis[u]==1) continue;
vis[u]=1;
for(int i=head[u];i;i=edge[i].next){
int v=edge[i].v;
int w=edge[i].w;
if(vis[v]==1) continue;
if(dis[u]+wcost[v]){
cost[v]=cost[u]+x[v];
q.push(v);
path[v]=u;
}
}
}
cout< st;
int tmp=b;
while(path[b]!=a){
st.push(path[b]);
b=path[b];
}
cout<
如果九月份再出这种类型的压轴题就完全可以pass了……