[GDKOI2016] Day1 不稳定的传送门 结论

对每个点可以走的传送门排序算期望即可
对于任意两个可以走的传送门推一个偏序关系就可以了

#include 
#include 
#include 
#include 

#define N 200050
#define INF 1<<30

using namespace std;
typedef double db;

db F[N];
int c[N],n,m;
struct Monster{int t,w; db p; }identity;
vector e[N];
inline void ut(db &x,db y) { x = min(x,y); }
bool cmp(Monster p1,Monster p2) {
    db v1 = (db)p1.w / p1.p + F[ p1.t ];
    db v2 = (db)p2.w / p2.p + F[ p2.t ];
    return v1 > v2;
}

void solve() {
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n-1;i++) scanf("%d",&c[i]);
    for (int i=0;i<=n+1;i++) F[i] = INF;
    F[n] = 0;

    for (int i=1;i<=n-1;i++) {
        Monster tmp = identity;
        tmp.t = i+1; tmp.p = 1; tmp.w = c[i];
        e[i].push_back(tmp);
    }
    for (int i=1;i<=m;i++) {
        Monster tmp = identity; int s;
        scanf("%d%d%lf%d",&s,&tmp.t,&tmp.p,&tmp.w);
        e[s].push_back(tmp);
    }

    for (int i=n-1;i>=1;i--) {
        sort(e[i].begin(),e[i].end(),cmp);
        for (int j=0;j<(int)e[i].size();j++) {
            int W = e[i][j].w;
            int T = e[i][j].t;
            db P = e[i][j].p;
            ut(F[i],W + P * F[T]+ ((db)1-P) * F[i]);
        }
    }

    printf("%.2lf\n",F[1]);
}

int main()
{
    freopen("portal.in","r",stdin);
    freopen("portal.out","w",stdout);
    solve();
    fclose(stdin); fclose(stdout);
    return 0;
}