【NOIP2016】换教室 题解
NOIP2016
核心思路
设dpi,j,0/1表示前i个教室，提交了j个申请，第i个教室是否成功的最小期望体力。
disi,j表示从i到j的最短距离，Floyd初始化。
下面是完整的转移方程。
dpi,j,0dpi,j,1=min{dpi−1,j,0+disci−1,cidpi−1,j,1+(1−ki−1)disci−1,ci+ki−1disdi−1,ci=min{dpi−1,j−1,0+disci−1,ci(1−ki)+disci−1,dikidpi−1,j−1,1+disdi−1,dikiki−1+disdi−1,ciki−1(1−ki)+disci−1,di(1−ki−1)ki+disci−1,ci(1−ki)(1−ki−1)
完整代码
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| #include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
#define int ll
#define REP(i,e,s) for(register int i=e; i<=s; i++)
#define DREP(i,e,s) for(register int i=e; i>=s; i--)
#define ll long long
#define DE(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG(a) DE("DEBUG: %d\n",a)
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
int read() {
int x=0,f=1,ch=getchar();
while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
const int MAXN=2000+10,INF=0x3f3f3f;
int c[MAXN],d[MAXN],a[MAXN][MAXN];
double dp[MAXN][MAXN][2],k[MAXN];
signed main() {
int n=read(),m=read(),v=read(),e=read();
REP(i,1,n) c[i]=read();
REP(i,1,n) d[i]=read();
REP(i,1,n) scanf("%lf",&k[i]);
REP(i,0,v) REP(j,0,v) a[i][j]=INF;
REP(i,1,v) a[i][i]=a[i][0]=a[0][i]=0;
REP(i,1,e) {
int x=read(),y=read(),w=read();
a[x][y]=a[y][x]=min(a[x][y],w);
}
REP(t,1,v) REP(i,1,v) REP(j,1,v) a[i][j]=min(a[i][j],a[i][t]+a[t][j]);
REP(i,0,n) REP(j,0,m) dp[i][j][0]=dp[i][j][1]=INF;
dp[1][0][0]=dp[1][1][1]=0;
DE("in");
REP(i,2,n) {
dp[i][0][0]=dp[i-1][0][0]+a[c[i-1]][c[i]];
REP(j,1,min(i,m)) {
dp[i][j][0]=min(dp[i-1][j][0]+a[c[i-1]][c[i]],dp[i-1][j][1]+a[c[i-1]][c[i]]*(1-k[i-1])+a[d[i-1]][c[i]]*k[i-1]);
dp[i][j][1]=min(dp[i-1][j-1][0]+a[c[i-1]][c[i]]*(1-k[i])+a[c[i-1]][d[i]]*k[i],
dp[i-1][j-1][1]+a[d[i-1]][d[i]]*k[i]*k[i-1]+a[d[i-1]][c[i]]*k[i-1]*(1-k[i])+a[c[i-1]][d[i]]*(1-k[i-1])*k[i]+a[c[i-1]][c[i]]*(1-k[i])*(1-k[i-1]));
}
}
double ans=INF;
REP(i,0,m) DE("%.2lf\n",ans=min(ans,min(dp[n][i][1],dp[n][i][0])));
printf("%.2lf\n",ans);
return 0;
}
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