Morgana is playing a game called End Fantasy VIX. In this game, characters have nn skills, every skill has its damage. And using skill has special condition. Briefly speaking, if this time you use skill "x", then next time you can use skill "y" (just like combo). There are mm conditions (xi, yiy_i), and you can't break the rules. (that means, if you don't have any condition that equals to (xx, yy), then you can't use "y" after use "x").
Now, Morgana wants to defeat the boss, he can use skills t times. In the first time he can use any skill, then he should obey the rules. Besides, he has a special armor called "Xue La", he can use this armor and add a debuff to the boss. The debuff will record damage and when it is over, the record damage will be caused again. (that means double damage) The debuff will continue TT times, and he can use this armor in any time, it won't be in conflict with skills.
Finally, Morgana wants to maximize the damage, but it is too difficult. So please help him deal with this problem.
(If Morgana can not use any skill at a time, he will finish the game and the final damage is his total damage at this time.)
Input
First line contains 44 integers n,m,t,T (2≤n≤642 le n le 64, 1≤m≤n×(n−1)1 le m le n imes (n-1) , 1≤t≤1e9, 1≤T≤t).
In the next mm lines each line contains two integers represent condition (xi,yix_i, y_i) (xi,yi≤nx_i, y_i le n) .
Then the next line contains nn integers represent the damage of the skills (every skill's damage is smaller than 1e81e8).
Output
One line with one integer.
思路
题意理解上有些歧义，从“If Morgana can not use any skill at a time, he will finish the game and the final damage is his total damage at this time.”这句理解的话应该是必须满T次才可以翻倍的，但据说场上有clarification说不满T次也可以结算。这篇题解是按照“必须满T次”写的，如果要改成不满T次也可以结算的话就没有必要对矩阵乘法分两种情况讨论了。
最大伤害可能有两种情况，即使用血棘(là)或不使用血辣。使用血辣的情况相当于在一个有向图上选择一条经过点数恰好为T的路径， 将路径上的点权和翻倍，然后在这条路径的首尾加上总数不超过(t-T)的点，使得总点权和最大。不使用血辣的情况则比较简单，直接选择一条经过点数不超过t的路径使得点权和最大即可。
联想到通过邻接矩阵乘法计算有向图上从u到v长度为T的路径条数的思路，不妨尝试将这里的问题转化成可以在矩阵上计算的问题。
用矩阵$A_{ij}$来表示图上从i到j的某些路径的点权和的最大值，如果路径不存在则定义为0。
用“路径合并”运算（即当一条路径的终点与另一条路径的起点均为k时，定义合并的结果为两条路径合并后的点权和）和$max$运算重定义矩阵乘法：$(A cdot B)_{ij} = max_{k}{A_{ik} mathop{Merge} B_{kj}}$.
由于$max$运算可交换、可结合、存在单位元0（由于题目中点权均为正数），路径合并运算可结合，且$max$对“路径合并”满足分配律（即$max(A_{ik}, B_{ik}) mathop{Merge} C_{kj} = max(A_{ik} mathop{Merge} C_{kj},B_{ik} mathop{Merge} C_{kj})$），可知重定义后的矩阵乘法是可结合的，即可以用快速幂的思路进行分治计算。
考虑上述运算的实际意义，如果我们将题目给出的有向图写成具有上述性质的矩阵$G$，则$G^{T-1}_{ij}$即为从i到j恰好经过T-1条边（即T个点）的所有路径的最大点权和。这样我们就知道，想要用血辣的话，只要对$G^{T-1}$中的每个元素翻倍就可以了。
最后的问题就是如何在这段使用血辣的路径前后加上总数不超过$(t-T)$的点使得路径长度最大。仍然是用快速幂的思路，我们只要考虑在(t-T)个(G+I)的连乘之间任选一个位置插入刚才翻倍后的$G^{T-1}$，把所有情况用$max$合并起来即可。
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 64; 5 int N, M, t, T, v[maxn]; 6 struct Mat 7 { 8 LL A[maxn][maxn]; 9 void Print() const 10 { 11 for(int i = 0;i < N;++i) 12 for(int j = 0;j < N;++j) printf("%lld%c", A[i][j], " "[j+1 == N]); 13 puts(""); 14 } 15 }; 16 17 bool tp;//因为懒得把运算符重载改成函数所以用了一个全局变量，一般认为这样写是不严谨的 18 //tp=1时表示路径长度可以小于t 19 Mat operator + (const Mat &a, const Mat &b) 20 { 21 Mat ans; 22 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 23 ans.A[i][j] = max(a.A[i][j], b.A[i][j]); 24 return ans; 25 } 26 27 Mat operator * (const Mat &a, const Mat &b) 28 { 29 Mat ans; 30 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 31 { 32 if(tp) ans.A[i][j] = max(a.A[i][j], b.A[i][j]); 33 else ans.A[i][j] = 0; 34 for(int k = 0;k < N;++k) 35 { 36 if(a.A[i][k] && b.A[k][j]) 37 ans.A[i][j] = max(ans.A[i][j], a.A[i][k] + b.A[k][j] - v[k]); 38 } 39 } 40 return ans; 41 } 42 43 Mat G, I; 44 void init() 45 { 46 scanf("%d%d%d%d", &N, &M, &t, &T); 47 int x, y; 48 while(M--) 49 { 50 scanf("%d%d", &x, &y); 51 --x, --y; 52 G.A[x][y] = 1; 53 } 54 for(int i = 0;i < N;++i) scanf("%d", &v[i]); 55 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 56 { 57 if(G.A[i][j] == 1) G.A[i][j] = v[i] + v[j]; 58 } 59 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 60 I.A[i][j] = 0; 61 for(int i = 0;i < N;++i) 62 I.A[i][i] = v[i]; 63 } 64 65 Mat powmod(Mat a, int n) 66 { 67 Mat ans = I; 68 while(n) 69 { 70 if(n & 1) ans = ans * a; 71 a = a * a; 72 n >>= 1; 73 } 74 return ans; 75 } 76 77 Mat powmod2(Mat a, Mat g, int n) 78 { 79 Mat ans = a, pw = I; 80 a = a * g + g * a; 81 while(n) 82 { 83 if(n & 1) 84 { 85 ans = ans + pw * a + a * pw; 86 pw = pw * g + g * pw; 87 } 88 n >>= 1; 89 a = a * g + g * a; 90 g = g * g; 91 } 92 return ans; 93 } 94 95 void work() 96 { 97 tp = false;//必须够T次 98 Mat a = powmod(G, T-1); 99 bool useXL = false; 100 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 101 if(a.A[i][j]) 102 { 103 useXL = true; 104 a.A[i][j] <<= 1; 105 } 106 LL ans = 0; 107 tp = true;//可以不足t次 108 if(useXL) 109 { 110 a = powmod2(a, G, t - T); 111 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) ans = max(ans, a.A[i][j]); 112 } 113 114 a = powmod(G, t); 115 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) ans = max(ans, a.A[i][j]); 116 printf("%lld ", ans); 117 } 118 119 int main() 120 { 121 init(); 122 work(); 123 return 0; 124 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 64; 5 int N, M, t, T, v[maxn]; 6 struct Mat 7 { 8 LL A[maxn][maxn]; 9 void Print() const 10 { 11 for(int i = 0;i < N;++i) 12 for(int j = 0;j < N;++j) printf("%lld%c", A[i][j], " "[j+1 == N]); 13 puts(""); 14 } 15 }; 16 17 Mat operator + (const Mat &a, const Mat &b) 18 { 19 Mat ans; 20 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 21 ans.A[i][j] = max(a.A[i][j], b.A[i][j]); 22 return ans; 23 } 24 25 Mat operator * (const Mat &a, const Mat &b) 26 { 27 Mat ans; 28 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 29 { 30 ans.A[i][j] = max(a.A[i][j], b.A[i][j]); 31 for(int k = 0;k < N;++k) 32 { 33 if(a.A[i][k] && b.A[k][j]) 34 ans.A[i][j] = max(ans.A[i][j], a.A[i][k] + b.A[k][j] - v[k]); 35 } 36 } 37 return ans; 38 } 39 40 Mat G, I; 41 void init() 42 { 43 scanf("%d%d%d%d", &N, &M, &t, &T); 44 int x, y; 45 while(M--) 46 { 47 scanf("%d%d", &x, &y); 48 --x, --y; 49 G.A[x][y] = 1; 50 } 51 for(int i = 0;i < N;++i) scanf("%d", &v[i]); 52 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 53 { 54 if(G.A[i][j] == 1) G.A[i][j] = v[i] + v[j]; 55 } 56 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 57 I.A[i][j] = 0; 58 for(int i = 0;i < N;++i) 59 I.A[i][i] = v[i]; 60 } 61 62 Mat powmod(Mat a, int n) 63 { 64 Mat ans = I; 65 while(n) 66 { 67 if(n & 1) ans = ans * a; 68 a = a * a; 69 n >>= 1; 70 } 71 return ans; 72 } 73 74 Mat powmod2(Mat a, Mat g, int n) 75 { 76 Mat ans = a, pw = I; 77 a = a * g + g * a; 78 while(n) 79 { 80 if(n & 1) 81 { 82 ans = ans + pw * a + a * pw; 83 pw = pw * g + g * pw; 84 } 85 n >>= 1; 86 a = a * g + g * a; 87 g = g * g; 88 } 89 return ans; 90 } 91 92 void work() 93 { 94 Mat a = powmod(G, T-1); 95 bool useXL = false; 96 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) 97 if(a.A[i][j]) 98 { 99 useXL = true; 100 a.A[i][j] <<= 1; 101 } 102 LL ans = 0; 103 if(useXL) 104 { 105 a = powmod2(a, G, t - T); 106 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) ans = max(ans, a.A[i][j]); 107 } 108 109 a = powmod(G, t); 110 for(int i = 0;i < N;++i) for(int j = 0;j < N;++j) ans = max(ans, a.A[i][j]); 111 printf("%lld ", ans); 112 } 113 114 int main() 115 { 116 init(); 117 work(); 118 return 0; 119 }