上学_ssl2404_dp

没有文字，放图

题解

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显而易见的dp
f[i][j]表示时刻i走到j点的最小花费，那么显而易见的
对于一张从i通往v，x时刻出发y时刻到达花费为cost的车票有f[v][y]=min(f[i][j]+cost,f[v][y])
不走，则f[i+1][j]=min(f[i][j],f[i+1][j])
答案就是f[t+1][n−1]

Code

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#include <stdio.h>#define inf 0x7fffffffusing namespace std;struct ticket{    int st,ed,lTime,aTime,cost;}v[51];int f[10001][51];int min(int x,int y){    return x<y?x:y;}int main(){    int n,t,m;    scanf("%d%d%d",&n,&t,&m);    for (int i=1;i<=m;i++)    {        scanf("%d%d%d%d%d",&v[i].st,&v[i].ed,&v[i].lTime,&v[i].aTime,&v[i].cost);        v[i].aTime+=v[i].lTime+1;    }    for (int i=0;i<=t+1;i++)        for (int j=0;j<n;j++)            f[i][j]=inf;    f[0][0]=0;    for (int i=0;i<=t;i++)        for (int j=0;j<n;j++)        {            if (f[i][j]!=inf)                for (int k=1;k<=m;k++)                    if (v[k].st==j&&v[k].lTime==i&&v[k].aTime<=t+1)                        f[v[k].aTime][v[k].ed]=min(v[k].cost+f[i][j],f[v[k].aTime][v[k].ed]);            f[i+1][j]=min(f[i][j],f[i+1][j]);        }    if (f[t+1][n-1]==inf)        f[t+1][n-1]=-1;    printf("%d\n",f[t+1][n-1]);    return 0;}