UVA-1336 Fixing the Great Wall (interval dp)

The main idea: the Great Wall (regarded as the x positive half axis) has the N place breakage. There is an intelligent repair robot that is known for its initial position and movement speed. Each broken place has a set of parameters (X,c,d), X is the position, C, D means that after the time t repair the damage at the cost of D*t+c. Use a robot to repair the minimum cost of all breakage.

Title Analysis: To the ultimate cost, you can not jump repair, that is, after a period of time has been repaired the damage should be a continuous interval. The definition of DP (I,J,K) indicates the cost of the robot staying at K (0 for the left end, 1 for the right end) after the repair (I,J). The cost of repairing a broken place is not fixed, but it grows linearly with time, so when one or a piece of damage is repaired, the cost of repairing the other breakage can be calculated by adding it to the current state, or as a time-cost for repairing a damaged place. State transfer equation: DP (i,j,1) =min (DP (i,j-1,0) +W1,DP (i,j-1,1) +w2) DP (i,j,0) =min (DP (i+1,j,0) +W3,DP (i+1,j,1) +w4) where W1, W2, W3, W4 for the corresponding time cost and repair cost.

The code is as follows:

# include<iostream># include<cstdio># include<cmath># include<cstring># include<    algorithm>using namespace Std;const int n=1005;const double inf=1e30;struct p{int x,c,dlt;    BOOL operator < (const P &a) const{return x<a.x; }};    P p[n];int n,v,x;double dp[n][n][2],s[n];d ouble dfs (int l,int r,int k) {if (dp[l][r][k]>-1.0) return dp[l][r][k];        if (l==r) {double T=fabs (double) x (double) p[l].x)/(double) v;        DP[L][R][K]=S[N]*T+P[L].C;    return dp[l][r][k];        } if (k==0) {double A=dfs (l+1,r,0);        Double B=dfs (l+1,r,1);        Double t1= (double) (p[l+1].x-p[l].x)/(double) v;        Double t2= (double) (p[r].x-p[l].x)/(double) v;        Double D=s[l]+s[n]-s[r];    Dp[l][r][k]=min (A+D*T1,B+D*T2) + (double) p[l].c;        }else{double A=dfs (l,r-1,0);        Double B=dfs (l,r-1,1);        Double t1= (double) (p[r].x-p[l].x)/(double) v;        Double t2= (double) (p[r].x-p[r-1].x)/(double) v; Double D=S[L-1]+S[N]-S[R-1];    Dp[l][r][k]=min (A+D*T1,B+D*T2) +p[r].c; } return dp[l][r][k];}        void Look () {for (int. i=1;i<=n;++i) {for (int j=1;j<=n;++j) printf ("%.lf", dp[i][j][0]);    cout<<endl;    } cout<<endl;        for (int i=1;i<=n;++i) {for (int j=1;j<=n;++j) printf ("%.lf", dp[i][j][1]);    cout<<endl;    }}int Main () {//freopen ("UVA-1336 Fixing the Great Wall.txt", "R", stdin);        while (~SCANF ("%d%d%d", &n,&v,&x)) {if (n+v+x==0) break;        for (int i=1;i<=n;++i) scanf ("%d%d%d", &AMP;P[I].X,&AMP;P[I].C,&AMP;P[I].DLT);        Sort (p+1,p+n+1);        s[0]=0.0;        for (int i=1;i<=n;++i) s[i]=s[i-1]+ (double) p[i].dlt;        Memset (Dp,-1.0,sizeof (DP));        printf ("%d\n", (int) min (dfs (1,n,0), DFS (1,n,1)));    Look (); } return 0;}

　　

UVA-1336 Fixing the Great Wall (interval dp)