HDU-5115 Dire Wolf(区间dp)

题意：有N匹狼排成一排，第i匹狼具有一个基础攻击力和对邻近单位的加成攻击力。攻击并杀死一匹狼受到的伤害 = 该狼自身的攻击力 + 邻近狼给它的加成攻击力。问你杀死这N匹狼受到的最少伤害。

你可以任选顺序杀死这N匹狼。当有a b c三匹狼时，杀死狼b，则狼a和狼c相邻。

思路：区间dp。dp[i][j]表示杀死[i, j]区间内所有狼承受的最小伤害。所以在第三层循环中枚举的是在区间[i, j]中杀死的最后一匹狼。

状态转移方程可以有两种写法：

① 杀最后一匹狼之前可以先杀前半段，也可以先杀后半段，只需替换相对应的加成伤害即可。

dp[i][j] = min(dp[i][j], dp[i][k]+dp[k+1][j]-extra[k+1]+extra[j+1]);
dp[i][j] = min(dp[i][j], dp[i][k]+dp[k+1][j]-extra[k]+extra[i-1]);

② 杀光前半段和后半段之后再杀最后一匹狼。

dp[i][j] = min(dp[i][j], dp[k+1][j]+fact[k]+extra[i-1]+extra[j+1]); (k == i)

dp[i][j] = min(dp[i][j], dp[i][k-1]+dp[k+1][j]+fact[k]+extra[i-1]+extra[j+1]) (i < k < j)

dp[i][j] = min(dp[i][j], dp[i][k-1]+dp[k+1][j]+fact[k]+extra[i-1]+extra[j+1]); (k == j)

两种转移方程异曲同工，实际上是一样的，只不过是在人文语言中的解释不同罢了。

Code1：

#include <bits/stdc++.h>using namespace std;const int maxn = 205;int fact[maxn], extra[maxn];int dp[maxn][maxn];int main(){int t, n;scanf("%d", &t);for(int _ = 1; _ <= t; ++_){scanf("%d", &n);extra[0] = extra[n+1] = 0;for(int i = 1; i <= n; ++i) scanf("%d", &fact[i]);for(int i = 1; i <= n; ++i) scanf("%d", &extra[i]);memset(dp, 0x3f, sizeof dp);for(int i = 1; i <= n; ++i) dp[i][i] = fact[i]+extra[i-1]+extra[i+1];for(int len = 2; len <= n; ++len)for(int i = 1; i <= n-len+1; ++i){int j = i+len-1;for(int k = i; k < j; ++k){dp[i][j] = min(dp[i][j], dp[i][k]+dp[k+1][j]-extra[k+1]+extra[j+1]);dp[i][j] = min(dp[i][j], dp[i][k]+dp[k+1][j]-extra[k]+extra[i-1]);}}printf("Case #%d: %d\n", _, dp[1][n]);}return 0;}

Code2：

#include <bits/stdc++.h>using namespace std;const int maxn = 205;int fact[maxn], extra[maxn];int dp[maxn][maxn];int main(){int t, n;scanf("%d", &t);for(int _ = 1; _ <= t; ++_){scanf("%d", &n);extra[0] = extra[n+1] = 0;for(int i = 1; i <= n; ++i) scanf("%d", &fact[i]);for(int i = 1; i <= n; ++i) scanf("%d", &extra[i]);memset(dp, 0x3f, sizeof dp);for(int i = 1; i <= n; ++i) dp[i][i] = fact[i]+extra[i-1]+extra[i+1];for(int len = 2; len <= n; ++len)for(int i = 1; i <= n-len+1; ++i){int j = i+len-1;for(int k = i; k <= j; ++k){if(k == i) dp[i][j] = min(dp[i][j], dp[k+1][j]+fact[k]+extra[i-1]+extra[j+1]);else if(k == j) dp[i][j] = min(dp[i][j], dp[i][k-1]+fact[k]+extra[i-1]+extra[j+1]);else dp[i][j] = min(dp[i][j], dp[i][k-1]+dp[k+1][j]+fact[k]+extra[i-1]+extra[j+1]);}}printf("Case #%d: %d\n", _, dp[1][n]);}return 0;}

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