标签:priority col string continue 赋值 ++i gis Fix def
颓了、、、重边导致我乖乖用邻接矩阵。。。。
好吧就是个最短路计数。。。。如果更新时d[v]==d[u]+w[i],就可以接起来,把两个加在一起。。
如果d[v]>d[u]+w[i],那么c[v]直接赋值为c[u],相当于这个最短路是由u转移过来的、
#include<cstdio> #include<iostream> #include<cstring> #include<queue> #define R register int using namespace std; const int N=2010,M=4000010; inline int g() { R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch==‘-‘?-1:fix; do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix; } int n,m,cnt; int w[N][N],fir[N],d[N],c[N]; bool vis[N]; priority_queue<pair<int,int> > q; inline void dijk() { memset(d,0x3f,sizeof(int)*(n+2)); c[1]=1,d[1]=0,q.push(make_pair(0,1)); while(q.size()) { R u=q.top().second; q.pop(); if(vis[u]) continue; vis[u]=true; for(R i=1;i<=n;++i) { if(d[i]>d[u]+w[u][i]) d[i]=d[u]+w[u][i],q.push(make_pair(-d[i],i)),c[i]=c[u]; else if(d[i]==d[u]+w[u][i]) c[i]+=c[u]; } } } signed main() { freopen("in.in","r",stdin); n=g(),m=g(); memset(w,0x3f,sizeof(w)); for(R i=1,u,v,ww;i<=m;++i) u=g(),v=g(),ww=g(),w[u][v]=min(w[u][v],ww); dijk(); if(d[n]==0x3f3f3f3f) printf("No answer\n"); else printf("%d %d\n",d[n],c[n]); }
2019.04.24
标签:priority col string continue 赋值 ++i gis Fix def
原文地址:https://www.cnblogs.com/Jackpei/p/10765789.html