BZOJ 2118 Dijkstra

思路：

经典题 不解释

找到最小的数mn

所有都是在mod mn的意义下 搞得

i->(i+a[i])%mn  边权为a[i]

//By SiriusRen
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N=*,inf=0x3f3f3f3f;
int n,first[N],next[N],v[N],tot,minn=inf,a[],vis[N];
ll BMin,BMax,w[N],dis[N],ans;
void add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;}
struct Node{int now,dis;}st,jy;
bool operator<(Node a,Node b){return a.dis>b.dis;}
priority_queue<Node>pq;
int main(){
    memset(first,-,sizeof(first));
    memset(dis,0x3f,sizeof(dis));
    scanf("%d%lld%lld",&n,&BMin,&BMax);
    for(int i=;i<=n;i++)scanf("%d",&a[i]),minn=min(minn,a[i]);
    dis[]=;
    for(int i=;i<=n;i++)
        for(int j=;j<minn;j++)
            add(j,(j+a[i])%minn,a[i]);
    pq.push(st);
    while(!pq.empty()){
        Node t=pq.top();pq.pop();
        if(vis[t.now])continue;
        vis[t.now]=;
        for(int i=first[t.now];~i;i=next[i])
            if(dis[v[i]]>dis[t.now]+w[i])
                dis[v[i]]=dis[t.now]+w[i],
                jy.now=v[i],jy.dis=dis[v[i]],pq.push(jy);
    }BMin--;
    for(int i=;i<minn;i++){
        if(dis[i]<=BMin)ans-=(BMin-dis[i])/minn+;
        if(dis[i]<=BMax)ans+=(BMax-dis[i])/minn+;
    }printf("%lld\n",ans);
}