题目大意:求s到t 长度为n 的最短路
裸的倍增Floyd
#include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> using namespace std; inline char nc() { static char buf[100000],*p1=buf,*p2=buf; if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; } return *p1++; } inline void read(int &x) { char c=nc(),b=1; for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1; for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b; } const int N=205; struct edge{ int u,v,w; }G[N]; int n,m,S,T; int f[N][N],h[N][N],tmp[N][N]; int sx[N],icnt; inline int Bin(int x){ return lower_bound(sx+1,sx+icnt+1,x)-sx; } int main() { freopen("t.in","r",stdin); freopen("t.out","w",stdout); read(n); read(m); read(S); read(T); for (int i=1;i<=m;i++) read(G[i].w),read(G[i].u),read(G[i].v),sx[++icnt]=G[i].u,sx[++icnt]=G[i].v; sort(sx+1,sx+icnt+1); icnt=unique(sx+1,sx+icnt+1)-sx-1; for (int i=1;i<=m;i++) G[i].u=Bin(G[i].u),G[i].v=Bin(G[i].v); memset(f,0x3f,sizeof(f)); memset(h,0x3f,sizeof(h)); for (int i=1;i<=icnt;i++) h[i][i]=0; for (int i=1;i<=m;i++) f[G[i].u][G[i].v]=f[G[i].v][G[i].u]=G[i].w; for (int t=0;(1<<t)<=n;t++) { if (n&(1<<t)) { memset(tmp,0x3f,sizeof(tmp)); for (int k=1;k<=icnt;k++) for (int i=1;i<=icnt;i++) for (int j=1;j<=icnt;j++) tmp[i][j]=min(tmp[i][j],h[i][k]+f[k][j]); memcpy(h,tmp,sizeof(tmp)); } memset(tmp,0x3f,sizeof(tmp)); for (int k=1;k<=icnt;k++) for (int i=1;i<=icnt;i++) for (int j=1;j<=icnt;j++) tmp[i][j]=min(tmp[i][j],f[i][k]+f[k][j]); memcpy(f,tmp,sizeof(tmp)); } printf("%d\n",h[Bin(S)][Bin(T)]); return 0; }