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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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