BZOJ 2118 墨墨的等式

模意义下最短路

挂题解
http://blog.csdn.net/PoPoQQQ/article/details/46605701

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define N 500005#define M 14using namespace std;namespace runzhe2000{    typedef long long ll;    const ll INF = 1ll<<60;    ll dis[N], Bmin, Bmax; bool inq[N];    int ecnt, n, last[N], a[M];    struct edge{int next, to, val;}e[N*M];    void addedge(int a, int b, int c)    {        e[++ecnt] = (edge){last[a], b, c};        last[a] = ecnt;    }    void SPFA()    {        queue<int> q; q.push(0);        for(int i = 0; i < a[1]; i++) dis[i] = INF;        dis[0] = 0;        for(; !q.empty(); q.pop())        {            int x = q.front(); inq[x] = 0;            for(int i = last[x]; i; i = e[i].next)            {                int y = e[i].to;                if(dis[x] + e[i].val < dis[y])                {                    dis[y] = dis[x] + e[i].val;                    if(!inq[y])q.push(y), inq[y] = 1;                }            }        }    }    ll calc(ll lim, ll a, int b) // 干，没注意到a也要看long long，调了好一会儿     {        if(lim < a) return 0;        else return (lim - a) / b + 1;    }    void main()    {        scanf("%d%lld%lld",&n,&Bmin,&Bmax);        for(int i = 1; i <= n; i++)        {            scanf("%d",&a[i]);            if(!a[i]){i--; n--; continue;}        }        sort(a+1, a+1+n);        for(int i = 0; i < a[1]; i++)            for(int j = 2; j <= n; j++)                addedge(i, (i+a[j])%a[1], a[j]);        SPFA();        ll ans = 0;        for(int i = 0; i < a[1]; i++)        {            if(dis[i] == INF) continue;            ans += calc(Bmax, dis[i], a[1]);            ans -= calc(Bmin-1, dis[i], a[1]);        }        printf("%lld\n",ans);    }}int main(){    runzhe2000::main();}