直接上代碼吧 Debug一下午 心累。。今天讓我對memset有了完整的認識

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
template<class T> inline void read(T &_a){
    bool f=0;int _ch=getchar();_a=0;
    while(_ch<‘0‘ || _ch>‘9‘){if(_ch==‘-‘)f=1
 
;_ch=getchar();}
    while(_ch>=‘0‘ && _ch<=‘9‘){_a=(_a<<1)+(_a<<3)+_ch-‘0‘;_ch=getchar();}
    if(f)_a=-_a;
}

const int maxn=2001,maxm=2001;
int n,m,v,e,dis[301][301],c[maxn],d[maxn];
double k[maxn],dp[maxn][maxn][2];

int main()
{
    read(n); read(m); read(v); read(e);
    
 
for (register int i=1;i<=n;++i) read(c[i]); 
    for (register int i=1;i<=n;++i) read(d[i]);
    for (register int i=1;i<=n;++i) scanf("%lf",&k[i]);
    memset(dis,0x3f,sizeof(dis));   // 如果是0x7f，那麽在之後的狀態轉移中會算爆int，直接GG
    for (register int i=1,a,b,w;i<=e;++i)
    {
        read(a); read(b); read(w);
        dis[b][a]
 
=dis[a][b]=min(dis[a][b],w);
    }
    for (register int i=0;i<=v;++i) dis[i][i]=dis[0][i]=0;
    for (register int i=1;i<=v;++i)
        for (register int j=1;j<=v;++j)
            for (register int k=1;k<=v;++k)
                if(i!=j&&j!=k&&k!=i)
                    dis[j][k]=min(dis[j][k],dis[j][i]+dis[i][k]);
    memset(dp,0x7f,sizeof(dp));
    dp[1][0][0]=dp[1][1][1]=0;
    for (register int i=2;i<=n;++i)
        for (register int j=0;j<=m&&j<=i;++j)
        {
            dp[i][j][0]=min(dp[i][j][0],dp[i-1][j][0]+dis[c[i-1]][c[i]]);
            dp[i][j][0]=min(dp[i][j][0],dp[i-1][j][1]+k[i-1]*dis[d[i-1]][c[i]]+(1.0-k[i-1])*dis[c[i-1]][c[i]]);
            if(j==0) continue;
            dp[i][j][1]=min(dp[i][j][1],dp[i-1][j-1][0]+k[i]*dis[c[i-1]][d[i]]+(1.0-k[i])*dis[c[i-1]][c[i]]);
            dp[i][j][1]=min(dp[i][j][1],dp[i-1][j-1][1]+k[i-1]*k[i]*dis[d[i-1]][d[i]]+k[i-1]*(1.0-k[i])*dis[d[i-1]][c[i]]+(1.0-k[i-1])*k[i]*dis[c[i-1]][d[i]]+(1.0-k[i-1])*(1.0-k[i])*dis[c[i-1]][c[i]]);
        }
    double ans=0x7fffffff;
    for (register int i=0;i<=m;++i) ans=min(ans,min(dp[n][i][0],dp[n][i][1]));
    printf("%.2lf",ans);
    return 0;
}

換教室 vijos2005 NOIP2016 D1T3 期望DP 圖論最短路（霧）