对照标答乱搞了一下1楼的程序就AC了QAQ

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int INF=1000000001;
const int maxn=105;
int d[maxn][maxn];
int n,m;
void printdis(int s,int t){
printf("%d\n",(d[s][t]==INF?-1:d[s][t]));
}
void addedge(int u,int v,int e){
if(d[u][v]<=e) return;
d[u][v]=e;
d[v][u]=e;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++){
d[i][j]=min(d[i][j],d[u][v]+d[i][u]+d[j][v]);
d[i][j]=min(d[i][j],d[u][v]+d[i][v]+d[j][u]);
d[j][i]=d[i][j];
}
}
int main(){
freopen("islet.in","r",stdin);
freopen("islet.out","w",stdout);
scanf("%d%d",&n,&m);
int num,s,t,u,v,e;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
d[i][j]=(i==j?0:INF);
while(m--){
scanf("%d",&num);
if(!num){
cin>>s>>t;
printdis(s,t);
}
else {
cin>>u>>v>>e;
addedge(u,v,e);
}
}
fclose(stdin);
fclose(stdout);
return 0;
}


这里是看起来很厉害的标答Q3Q 为什么要松弛2次QAQ

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int INF=1000000000;
const int maxn=105;
int n,m;
int d[maxn][maxn];
void init(){
cin>>n>>m;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
d[i][j]=INF;
for(int i=1;i<=n;i++) d[i][i]=0;
}
void relax(int x,int y,int len){
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++){
if(d[i][x]+d[y][j]+len<d[i][j]) d[i][j]=d[i][x]+d[y][j]+len;
if(d[i][y]+d[x][j]+len<d[i][j]) d[i][j]=d[i][y]+d[x][j]+len;
}
}
void solve(){
for(int i=1;i<=m;i++){
int o,x,y,len;
cin>>o;
if(!o){
cin>>x>>y;
if(d[x][y]==INF) cout<<"-1\n";
else cout<<d[x][y]<<"\n";
}
else{
cin>>x>>y>>len;
relax(x,y,len);
}
}
}
int main(){
// freopen("islet.in","r",stdin);
// freopen("islet.out","w",stdout);
init();
solve();
return 0;
}


floyd无脑乱搞O（n^3*m) QAQ

#include<cstdio>
#include<algorithm>
using namespace std;
const int INF=1000000001;
const int maxn=105;
int d[maxn][maxn];
int n,m;
int main(){
freopen("islet.in","r",stdin);
// freopen("islet.out","w",stdout);
scanf("%d%d",&n,&m);
int num,s,t,u,v,e;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
d[i][j]=(i==j?0:INF);
while(m--){
scanf("%d",&num);
if(!num){
scanf("%d%d",&s,&t);
printf("%d\n",(d[s][t]==INF?-1:d[s][t]));
}
if(num){
scanf("%d%d%d",&u,&v,&e);
d[u][v]=min(d[u][v],e);
d[v][u]=min(d[u][v],e);
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
d[i][j]=min(d[i][k]+d[k][j],d[i][j]);
}
}
fclose(stdin);
// fclose(stdout);
return 0;
}