题目思路：
dp[i][j][k]代表打到第i个怪物使用了第j种武器第k种属性,注意间接转化可能花的时间更少，所以要用弗洛伊德求武器和属性转化的最短路，然后在dp时武器和属性的状态转移要分开，不然会超时
#include<bits/stdc++.h> #define ll long long using namespace std; ll a,b,c,n,ans;// a is weapon, b is attribute, c is kind of monster, n is number of monster, ans is the answer ll dp[101][101][101]; ll x[101],y[101];// Time required to start using weapons and attributes ll f[101][101],g[101][101];// change time of weapons and attributes ll h[21][101][101]; // the impair of monster ll w[101],t[101]; // the blood and class of monster // *************maybe need use long long or double******************** void solve(); int main() { //输入 scanf("%lld%lld%lld%lld",&a,&b,&c,&n); for(int i = 1;i<=a;i++) scanf("%lld",&x[i]); for(int i = 1;i<=b;i++) scanf("%lld",&y[i]); for(int i = 1;i<=a;i++) for(int j = 1;j<=a;j++) scanf("%lld",&f[i][j]); for(int i = 1;i<=b;i++) for(int j = 1;j<=b;j++) scanf("%lld",&g[i][j]); for(int i = 1;i<=c;i++) for(int j = 1;j<=a;j++) for(int k = 1;k<=b;k++) scanf("%lld",&h[i][j][k]); for(int i = 1;i<=n;i++) scanf("%lld",&w[i]); for(int i = 1;i<=n;i++) scanf("%lld",&t[i]); solve(); printf("%lld",ans); return 0; } void solve() { //间接转化可能时间更短 ，所以求最短路 for(int k=1; k<=a; k++) for(int i=1; i<=a; i++) for(int j=1; j<=a; j++) f[i][j] = min(f[i][j],f[i][k]+f[k][j]); for(int k=1; k<=b; k++) for(int i=1; i<=b; i++) for(int j=1; j<=b; j++) g[i][j] = min(g[i][j],g[i][k]+g[k][j]); //初始化 memset(dp,0x3f,sizeof(dp)); for(int i=1; i<=a; i++) for(int j=1; j<=b; j++) dp[0][i][j]=x[i]+y[j]; //dp代表打到第i个怪物使用了第j种武器第k种属性 ，武器属性分开加 for(int i=1; i<=n; i++) { for(int j=1; j<=a; j++) for(int k=1; k<=b; k++) for(int p=1; p<=a; p++) dp[i][j][k] = min(dp[i][j][k],dp[i-1][p][k]+f[p][j]); for(int j=1; j<=a; j++) for(int k=1; k<=b; k++) for(int p=1; p<=b; p++) dp[i][j][k] = min(dp[i][j][k],dp[i][j][p]+g[p][k]); for(int j=1; j<=a; j++) for(int k=1; k<=b; k++) dp[i][j][k]+=w[i]/h[t[i]][j][k]+(w[i]%h[t[i]][j][k]!=0); } ans = 1e18;//不能0x3f3f3f3f for(int i = 1;i<=a;i++) for(int j = 1;j<=b;j++) ans = min(ans,dp[n][i][j]); }