换教室
-
f[i][j][0/1]表示前i个时间点,共申请了j次,第i个时间点否/是进行了申请。
-
dis[a][b]表示a教室->b教室的距离
-
c[i]表示默认的教室
-
d[i]表示更换后的教室
-
k[i]表示第i个教室申请成功的概率
【解题思路】
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> #include<queue> using namespace std; const int inf = 0x3fffffff; double k[2005],f[2005][2005][2]; int n,m,v,e,dis[305][305],d[2005],c[2005]; void work(){ for(int i = 1; i <= n; i++) for(int j = 0; j <= m; j++) f[i][j][0] = f[i][j][1] = inf; f[0][0][0] = 0;
//可以写成:
//f[1][0][0] = 0; f[1][1][1] = 0; for(int i = 1; i <= n; i++) for(int j = 0; j <= m; j++){ f[i][j][0] = min(f[i-1][j][0] + dis[c[i-1]][c[i]], f[i-1][j][1] + k[i-1]*(dis[d[i-1]][c[i]]) + (1-k[i-1])*dis[c[i-1]][c[i]]);//error:d[i-1]->d[i] if(j != 0) f[i][j][1] = min(f[i-1][j-1][0] + (1-k[i])*dis[c[i-1]][c[i]] + k[i]*dis[c[i-1]][d[i]],f[i-1][j-1][1] + k[i]*k[i-1]*dis[d[i-1]][d[i]] + (1-k[i-1])*(1-k[i])*dis[c[i-1]][c[i]] + k[i-1]*(1-k[i])*dis[d[i-1]][c[i]] + (1-k[i-1])*k[i]*dis[c[i-1]][d[i]]); } } int main(){ scanf("%d%d%d%d",&n,&m,&v,&e); for(int i = 1; i <= n; i++) scanf("%d",&c[i]); for(int i = 1; i <= n; i++) scanf("%d",&d[i]); for(int i = 1; i <= n; i++) scanf("%lf",&k[i]); for (int i=1;i<=v;i++) for (int j=1;j<i;j++) dis[j][i]=dis[i][j]=inf; for (int i=1;i<=e;i++){ int a,b,w; scanf("%d%d%d",&a,&b,&w); dis[a][b]=min(dis[a][b],w);// dis[b][a]=dis[a][b]; } for(int k = 1; k <= v; k++) for(int i = 1; i <= v; i++) for(int j = 1; j < i; j++) if(dis[i][k] + dis[k][j] < dis[i][j]) dis[i][j] = dis[j][i] = dis[i][k] + dis[k][j] ; work(); double ans = inf; for(int i = 0; i <= m; i++)//error:i=0 -> i = 1 for(int j = 0; j <= 1; j++) ans = min(ans, f[n][i][j]); printf("%0.2lf",ans); return 0; }
绿豆蛙的归宿
待定