HDU 5115 Dire Wolf

The last Beijing match on the scene, yesterday did a section DP, suddenly think of this problem, are very similar, turned out to do

Just started like yesterday, the same way, thinking about how to reverse, and later found that the problem should be pushed

In fact, the forward and reverse push at first glance looks very similar, only one is dp[i][j] said I, J and other wolves to eliminate the cost of i-j this period, one is to say that the last remaining i-j when the cost of eliminating this paragraph

To summarize the similarities between the DP and today's interval DP, find that the interval DP is applicable to the cost over time, or that DP is changing

#include <stdio.h> #include <algorithm> #include <string.h>using namespace Std;const int inf= 999999999;int a[205],b[205];int dp[205][205],n;void Init () {for (int. i=0;i<205;i++) {for (int j=0;j<205;j++) {dp[i ][j]=inf;}} b[n+1]=0;for (int i=1;i<=n;i++) dp[i][i]=a[i]+b[i-1]+b[i+1];for (int i=1;i<n;i++) dp[i][i+1]=min (dp[i][i]+a[i+ 1]+B[I+2]+B[I-1],DP[I+1][I+1]+A[I]+B[I-1]+B[I+2]);} int main () {#ifndef online_judgefreopen ("In.txt", "R", stdin), #endifint t;scanf ("%d", &t); int Cas=1;while (t--) { scanf ("%d", &n), for (int i=1;i<=n;i++) {scanf ("%d", &a[i]);} for (int i=1;i<=n;i++) {scanf ("%d", &b[i]);} Init (); for (int. l=3;l<=n;l++) {for (int i=1;i+l-1<=n;i++) {int j=i+l-1;dp[i][j]=min (dp[i][j],dp[i+1][j]+a[i]+b[ I-1]+B[J+1]);DP [I][j]=min (Dp[i][j],dp[i][j-1]+a[j]+b[j+1]+b[i-1]); for (int k=i+1;k<=j-1;k++) {dp[i][j]=min (dp[ I][j],dp[i][k-1]+dp[k+1][j]+b[i-1]+b[j+1]+a[k]);} printf ("dp[%d][%d]:%d\n", I,j,dp[i][j]);}} printf ("Case #%d:%d\n", Cas++,dp[1][n]);}}

HDU 5115 Dire Wolf