HDU 5115 Dire Wolf "Interval DP"

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5115

Test instructions
There are n wolves, each wolf has two kinds of properties, a kind of attack value, we did not kill a wolf, then we received the damage value of the wolf's attack value with the two wolves next to the added value of the and, to kill all the wolves to the minimum damage value.

Code:

#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <ctime>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <assert.h>using namespace STD;intNinta[ -],b[ -],dp[ -][ -];intMain () {intT, cases =1;scanf("%d", &t); while(t--) {scanf("%d", &n); for(inti =0; I <= N; i++) for(intj =0; J <= N; J + +) Dp[i][j] =100000000;memsetA0,sizeof(a));memset(b,0,sizeof(b)); for(inti =1; I <= N; i++)scanf("%d", &a[i]); for(inti =1; I <= N; i++)scanf("%d", &b[i]); for(inti =1; I <= N; i++) Dp[i][i] = A[i] + b[i-1] + b[i +1]; for(intL =1; L <= N; l++) { for(inti =1; i + L <= N; i++) {intj = i + L; Dp[i][j] = min (dp[i+1][J] + b[i-1] + b[j+1] + a[i], dp[i][j-1] + A[j] + b[j +1] + b[i-1]); for(intK = i+1; K <= J; k++) {Dp[i][j] = min (dp[i][j], dp[i][k-1] + dp[k +1][J] + a[k] + b[i-1] + b[j +1]); }            }        }printf("Case #%d:%d\n", cases++, dp[1][n]); }return 0;}

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Hdu 5115 Dire Wolf "Interval dp"