这题数据规模并不大！！这是重点………
所以直接暴力DP就好了：f[i]表示前 i 天的最小花费，则有$f[i]=min\{f[j]+cost[j+1][i]+k\} (0\leq j \leq i-1)$其中cost数组表示第L天到第R天只用一种运输方案连续运$R-L+1$天的最小代价，因为n很小所以暴力枚举L、R，用最短路算法来算……
记得开long long
1 /************************************************************** 2 Problem: 1003 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:52 ms 7 Memory:1396 kb 8 ****************************************************************/ 9 10 //BZOJ 1003 11 #include<vector> 12 #include<cstdio> 13 #include<cstring> 14 #include<cstdlib> 15 #include<iostream> 16 #include<algorithm> 17 #define rep(i,n) for(int i=0;i<n;++i) 18 #define F(i,j,n) for(int i=j;i<=n;++i) 19 #define D(i,j,n) for(int i=j;i>=n;--i) 20 #define pb push_back 21 using namespace std; 22 inline int getint(){ 23 int v=0,sign=1; char ch=getchar(); 24 while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();} 25 while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();} 26 return v*sign; 27 } 28 const int N=110,M=1000,INF=100000000; 29 typedef long long LL; 30 /******************tamplate*********************/ 31 int to[M],next[M],head[N],len[M],cnt; 32 void ins(int x,int y,int z){ 33 to[++cnt]=y; next[cnt]=head[x]; head[x]=cnt; len[cnt]=z; 34 } 35 void add(int x,int y,int z){ 36 ins(x,y,z);ins(y,x,z); 37 } 38 /*******************edge************************/ 39 int n,m,k,e; 40 LL f[N],cost[N][N],d[N]; 41 int Q[M]; 42 bool can[N][N],vis[N],inq[N]; 43 LL spfa(){ 44 int l=0,r=-1; 45 F(i,2,m) d[i]=INF; 46 d[1]=0;inq[1]=1; 47 Q[++r]=1; 48 while(l<=r){ 49 int x=Q[l++]; 50 inq[x]=0; 51 for(int i=head[x];i;i=next[i]) 52 if(vis[to[i]] && d[to[i]]>d[x]+len[i]){ 53 d[to[i]]=d[x]+len[i]; 54 if(!inq[to[i]]){ 55 Q[++r]=to[i]; 56 inq[to[i]]=1; 57 } 58 } 59 } 60 return d[m]; 61 } 62 63 int main(){ 64 #ifndef ONLINE_JUDGE 65 freopen("1003.in","r",stdin); 66 freopen("1003.out","w",stdout); 67 #endif 68 n=getint(); m=getint(); k=getint(); e=getint(); 69 int x,y,z; 70 F(i,1,e){ 71 x=getint(); y=getint(); z=getint(); 72 add(x,y,z); 73 } 74 int q=getint(); 75 F(i,1,m) F(j,1,n) can[i][j]=1; 76 F(i,1,q){ 77 x=getint(); y=getint(); z=getint(); 78 F(j,y,z) can[x][j]=0; 79 } 80 F(i,1,n) F(j,i,n){ 81 F(x,1,m) vis[x]=1; 82 F(x,2,m-1) F(k,i,j) if(!can[x][k]) vis[x]=0; 83 cost[i][j]=spfa()*(j-i+1); 84 } 85 F(i,1,n) f[i]=cost[1][i]; 86 F(i,1,n) F(j,0,i-1) f[i]=min(f[i],f[j]+cost[j+1][i]+k); 87 printf("%lld\n",f[n]); 88 return 0; 89 } 90