Luogu P5122 [USACO18DEC]Fine Dining 最短路

先跑一遍n为起点最短路，再新开一个点，向有干草垛的点连一根边权为d[u]-w的有向边（很重要。。我当时连的无向边，然后我死了。），相当于用价值抵消一部分边权，

然后以这个新的点为起点跑最短路就好了。。。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#define pc(x) putchar(x)
#define R register int
using namespace std;
const int N=50010,M=100010;
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,k,cnt;
int vr[M<<1],nxt[M<<1],w[M<<1],fir[N],d[N],f[N];
bool vis[N];
priority_queue<pair<int,int> > q;
inline void add(int u,int v,int ww) {vr[++cnt]=v,w[cnt]=ww,nxt[cnt]=fir[u],fir[u]=cnt;}
inline void dijk() {
    memset(d,0x3f,sizeof(int)*(n+2)); d[n]=0,q.push(make_pair(0,n));
    while(q.size()) {
        R u=q.top().second; q.pop(); if(vis[u]) continue; vis[u]=true;
        for(R i=fir[u];i;i=nxt[i]) { R v=vr[i];
            if(d[v]>d[u]+w[i]) d[v]=d[u]+w[i],q.push(make_pair(-d[v],v));
        }
    }
}
inline void dijk2() {
    memset(f,0x3f,sizeof(int)*(n+3)); f[n+1]=0,q.push(make_pair(0,n+1));
    memset(vis,false,sizeof(bool)*(n+3)); while(q.size()) {
        R u=q.top().second; q.pop(); if(vis[u]) continue; vis[u]=true;
        for(R i=fir[u];i;i=nxt[i]) { R v=vr[i];
            if(f[v]>f[u]+w[i]) f[v]=f[u]+w[i],q.push(make_pair(-f[v],v));
        }
    }
}
signed main() {
    n=g(),m=g(),k=g(); for(R i=1,u,v,w;i<=m;++i) u=g(),v=g(),w=g(),add(u,v,w),add(v,u,w); 
    dijk(); for(R i=1,u,w;i<=k;++i) u=g(),w=g(),add(n+1,u,d[u]-w); dijk2();
    for(R i=1;i<n;++i) f[i]<=d[i]&&d[i]!=0x3f3f3f3f?(pc('1'),pc('\n')):(pc('0'),pc('\n'));
}

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2019.04.24