Bzoj 1571: [usaco 2009 Open] ski course ski

Description

Farmer John wants to ski with Bessie in Colorado. Unfortunately, Bessie's skiing technology is not superb. Bessie understands that S (0 <= S <= 100) Skiing Courses are offered every day at the ski resort. Lesson I starts with m_ I (1 <= m_ I <= 10000), and the last time is L_ I (1 <= L_ I <= 10000 ). After completing section I, Bessie's skiing capacity will change to a_ I (1 <= a_ I <= 100). Note: This capacity is absolute, not a growth value. Bessie bought a map showing N (1 <= n <= 10,000) slopes for skiing, duration d_ I (1 <= d_ I <= 10000) required to slide from the top of slope I to the bottom ), and the skiing capacity C_ I (1 <= C_ I <= 100) required for each slope to ensure the safety of skiing. Bessie's ability must be greater than or equal to this level so that she can slide securely. Bessie can spend her time skiing, attending classes, or drinking a cup of cocoa, but she must leave the ski resort at T (1 <= T <= 10000. This means that she had to complete the last skiing before t moment. Find the maximum number of skiing times that Bessie can complete in the implementation. At the beginning of the day, her skiing capacity was 1. Input

Row 3: Three integers separated by spaces: T, S, N.

2nd ~ Row S + 1: The ski class I + 1 is described by an integer separated by three spaces: m_ I, L_ I, a_ I.

S + 2 ~ S + n + 1 row:

Line S + I + 1 uses an integer separated by two spaces to describe the I slice: C_ I, d_ I. Output

An integer that indicates the maximum number of skiing times that Bessie can complete within the time limit.

Question:

For dynamic planning, F [I] [J] indicates the maximum number of skiing times with the capability value J During I time.

Corresponding to the last three options:

① Delicious drink a cup of cocoa juice f [I] [J] = f [I-1] [J],

② Class F [I] [J] = f [one time before class] [arbitrary],

③ Skiing f [I] [J] = f [I-po [J] [J] + 1 (po [J] is a one-time skiing event with a capacity value <= J shortest time ).

For ② the last start time of the course where ke [I] [J] can end at the I moment and the capability value reaches J. In the DP process, G [I] = max {f [I] [J]} is processed.

G [T] is the answer.

Code:

#include<cstdio>#include<cstring>#include<algorithm>//by zrt//problem:using namespace std;int f[10005][105];int t,s,n;int ke[10005][105];int po[105];int g[10005];int main(){    #ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    #endif    memset(f,128,sizeof f);    scanf("%d%d%d",&t,&s,&n);    for(int i=1,m,l,a;i<=s;i++){        scanf("%d%d%d",&m,&l,&a);        ke[m+l-1][a]=max(ke[m+l-1][a],m);    }    memset(po,0x3f,sizeof po);    for(int i=1,c,d;i<=n;i++){        scanf("%d%d",&c,&d);        for(int j=c;j<=100;j++){            po[j]=min(po[j],d);        }    }    f[0][1]=0;    g[0]=0;    for(int i=1;i<=t;i++){        for(int j=1;j<=100;j++){            f[i][j]=f[i-1][j];            if(ke[i-1][j])f[i][j]=max(f[i][j],g[ke[i-1][j]]);            if(i-po[j]>=0)f[i][j]=max(f[i][j],f[i-po[j]][j]+1);            g[i]=max(g[i],f[i][j]);        }    }    printf("%d\n",g[t]);    return 0;}

Bzoj 1571: [usaco 2009 Open] ski course ski