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为了解决这个问题，我们需要找到消灭所有狼的最小代价。每次消灭一只狼时，需要支付这只狼的a攻击力和它旁边的狼的b攻击力。消灭一只狼后，左右两边的狼会并在一起。我们需要找到最小代价。

方法思路

解决代码

#include 
   
    
using namespace std;
ll read() {
    ll c = getchar();
    ll Nig = 1;
    ll x = 0;
    while (!isdigit(c) && c != '-') {
        c = getchar();
    }
    if (c == '-') {
        Nig = -1;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (c - '0');
        c = getchar();
    }
    return Nig * x;
}
int main() {
    int T = read();
    for (; T--; ) {
        int n = read();
        vector
    
      a(n + 2), b(n + 2);
        for (int i = 1; i <= n; ++i) {
            a[i] = read();
            b[i] = read();
        }
        ll INF = 0x3f3f3f3f;
        vector
     
      
       > dp(n + 2, vector
       
        (n + 2, INF));
        for (int i = 1; i <= n; ++i) {
            dp[i][i] = a[i] + (i > 1 ? b[i - 1] : 0) + (i < n ? b[i + 1] : 0);
        }
        for (int i = n - 1; i >= 1; --i) {
            for (int j = i + 1; j <= n; ++j) {
                ll temp1 = dp[i + 1][j] + a[i] + (i > 1 ? b[i - 1] : 0) + (j < n ? b[j + 1] : 0);
                ll temp2 = dp[i][j - 1] + a[j] + (i > 1 ? b[i - 1] : 0) + (j < n ? b[j + 1] : 0);
                dp[i][j] = min(temp1, temp2);
            }
        }
        cout << dp[1][n] << endl;
    }
}
       
      
     
    
   

代码解释

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