BZOJ 1050: [HAOI2006]旅行comf (并查集 或 单调队列)

这是建空间后做的第一道题啊= =好水

排序，枚举最小边，然后并查集求出联通时的最大边

或者排次序，从小到大插边，如果插边时最小的边拿掉不会使s与t不联通，就删去。

code：

#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> using namespace std; struct node{     int x,y,dist; }a[5010]; int n,m,s,t,f[510],ansi,ansa; bool cmp(node x,node y){return x.dist<y.dist;} int find(int x){if (x!=f[x]) f[x]=find(f[x]);return (f[x]);} int gcd(int x,int y){     if (!y) return x;     return gcd(y,x %y); } int main(){     scanf("%d%d",&n,&m);     for (int i=1;i<=m;i++) scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].dist);     sort(a+1,a+1+m,cmp);     scanf("%d%d",&s,&t);     ansi=a[1].dist;ansa=40000;     for (int i=1;i<=n;i++) f[i]=i;     for (int i=1;i<=m;i++) {         int x=find(a[i].x),y=find(a[i].y);         if (x!=y) f[x]=y;         x=find(s);y=find(t);         if (x==y) {             ansa=a[i].dist;             break;         }     }     if (ansa==40000) {printf("IMPOSSIBLE ");return 0;}     for (int i=2;i<=m;i++) {         for (int j=1;j<=n;j++) f[j]=j;         for (int j=i;j<=m;j++) {             int x=find(a[j].x),y=find(a[j].y);             if (x!=y) f[x]=y;             x=find(s);y=find(t);             if (x==y) {                 if (ansa*1.0/ansi>a[j].dist*1.0/a[i].dist) {ansa=a[j].dist;ansi=a[i].dist;}                 break;             }         }     }     int t=gcd(ansa,ansi);     if (ansa*1.0/ansi==ansa/ansi) printf("%d ",ansa/ansi);     else printf("%d/%d",ansa/t,ansi/t);     return 0; }