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渐渐认识到区域赛更侧重的是思维及基本算法的灵活运用，而不是算法的量(仅个人见解)，接下来要更多侧重思维训练了。
区间DP,dp[i][j]表示从i到j最终剩余第i 与第j只的最小伤害值，设置0与n+1两个虚拟的狼，状态转移方程如下：
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + b[i] + b[j])
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 208, INF = 0x3F3F3F3F;
#define MS(a, num) memset(a, num, sizeof(a))
#define PB(A) push_back(A)
#define FOR(i, n) for(int i = 0; i < n; i++)
int a[N], b[N];
int dp[N][N];
int main(){
int t;
cin>>t;
for(int cas = 1; cas <= t; cas++){
int n;
scanf("%d", &n);
int sum = 0;
for(int i = 1; i <= n; i++){
scanf("%d", &a[i]);
sum += a[i];
}
for(int i = 1; i <= n; i++){
scanf("%d", &b[i]);
}
b[n + 1] = b[0] = 0;
memset(dp, 0x3F, sizeof(dp));
for(int i = 0 ; i <= n + 1; i++){
dp[i][i] = 0;
dp[i][i + 1] = 0;
}
for(int len = 3;len <= n + 2; len++){
for(int i = 0, j = len - 1; j <= n + 1; i++, j++){
for(int k = i + 1; k < j; k++){
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + b[i] + b[j]);
}
}
}
printf("Case #%d: %d\n", cas, dp[0][n + 1] + sum);
}
return 0;
}
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原文地址：http://www.cnblogs.com/IMGavin/p/5746808.html
