P.S.o(︶︿︶)o 唉~虽然这题方程不难,但题目长,代码长,我花了超过3小时!(>﹏<)悲伤辣么大~~~ 谨此题解惠及众人,hh。
题意:给定长度为M的一串颜色序列,和平面上的N个颜色隧道。要求以颜色序列的顺序通过颜色隧道。(隧道可多次使用,可交叉,互不相同。)问从源点到汇点依次通过颜色的最小距离。
解法:f[i][j]表示通过颜色序列前 i 个颜色,这次的第 i 个颜色的隧道选 j 的最小距离。枚举第 i-1 个颜色选的隧道为 k ,则:f[i][j] = min( f[i][j] , f[i-1][k] + dis( a[k].xx , a[k].yy , a[j].x , a[j].y) + a[j].l );
我是规定每个隧道从(x,y)到(xx,yy)通过,所以读入时存2*N个隧道,(x,y)和(xx,yy)调换。这样就可以减去大部分人分f[i][j][0] 和 f[i][j][1]的多情况分析讨论,具体优越处见代码。
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1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 #include<cmath> 7 using namespace std; 8 #define M 35 9 #define N 65 10 #define C 105 11 #define INF 2e9 12 13 int m,n; 14 double xs,ys,xt,yt; 15 int q[M]; 16 struct node{double x,y,xx,yy,l;int c;}; 17 node a[2*N]; 18 double f[2][2*N]; 19 20 void ins(int id,double x,double y,double xx,double yy,int c) 21 { 22 a[id].x=x,a[id].y=y,a[id].c=c; 23 a[id].xx=xx,a[id].yy=yy; 24 a[id].l=sqrt((x-xx)*(x-xx)+(y-yy)*(y-yy)); 25 } 26 double mmin(double x,double y) {return x<y?x:y;} 27 double dis(double x,double y,double xx,double yy) {return sqrt((x-xx)*(x-xx)+(y-yy)*(y-yy));} 28 29 int main() 30 { 31 int T,i,j,k; 32 scanf("%d",&T); 33 while (T--) 34 { 35 scanf("%lf%lf%lf%lf",&xs,&ys,&xt,&yt); 36 scanf("%d",&m); 37 for (i=1;i<=m;i++) scanf("%d",&q[i]); 38 scanf("%d",&n); 39 for (i=1;i<=n;i++) 40 { 41 double x,y,xx,yy;int c; 42 scanf("%lf%lf%lf%lf%d",&x,&y,&xx,&yy,&c); 43 ins(2*i-1,x,y,xx,yy,c),ins(2*i,xx,yy,x,y,c); 44 } 45 n*=2; 46 47 double ans=INF; 48 for (j=1;j<=n;j++) 49 { 50 if (a[j].c!=q[1]) continue; 51 f[1][j]=dis(xs,ys,a[j].x,a[j].y)+a[j].l; 52 if (m==1) ans=mmin(ans,f[1][j]+dis(a[j].xx,a[j].yy,xt,yt));//xx,yy 53 } 54 int u=1; 55 for (i=2;i<=m;i++) 56 { 57 u=1-u; 58 for (j=1;j<=n;j++) 59 { 60 f[u][j]=INF; 61 if (a[j].c!=q[i]) continue; 62 for (k=1;k<=n;k++) 63 { 64 if (a[k].c!=q[i-1]) continue; 65 f[u][j]=mmin(f[u][j],f[1-u][k]+dis(a[k].xx,a[k].yy,a[j].x,a[j].y)+a[j].l); 66 } 67 if (i==m) ans=mmin(ans,f[u][j]+dis(a[j].xx,a[j].yy,xt,yt));//xx,yy 68 } 69 } 70 printf("%.4lf ",ans); 71 } 72 return 0; 73 }
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1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<cmath> 6 using namespace std; 7 8 const int N=110,M=110; 9 const double INF=(double)1e9; 10 int n,m; 11 int c[N]; 12 double d[N][M][2]; 13 struct node{double x,y;}; 14 struct edge{node s,d;double len;int c;}a[N]; 15 double minn(double x,double y){return x<y ? x:y;} 16 double dist(node t1,node t2){ 17 return sqrt((t1.x-t2.x)*(t1.x-t2.x)+(t1.y-t2.y)*(t1.y-t2.y)); 18 } 19 20 int main() 21 { 22 int T; 23 scanf("%d",&T); 24 while(T--) 25 { 26 node st,ed; 27 scanf("%lf%lf%lf%lf",&st.x,&st.y,&ed.x,&ed.y); 28 scanf("%d",&n); 29 for(int i=1;i<=n;i++) 30 scanf("%d",&c[i]); 31 scanf("%d",&m); 32 for(int i=1;i<=m;i++) 33 { 34 scanf("%lf%lf%lf%lf%d",&a[i].s.x,&a[i].s.y,&a[i].d.x,&a[i].d.y,&a[i].c); 35 a[i].len=dist(a[i].s,a[i].d); 36 } 37 double mn=INF; 38 for(int i=1;i<=m;i++) 39 { 40 if(a[i].c==c[1]) 41 { 42 d[1][i][0]=dist(st,a[i].s)+a[i].len; 43 d[1][i][1]=dist(st,a[i].d)+a[i].len; 44 if(n==1) mn=minn(mn,minn(d[1][i][0]+dist(a[i].d,ed),d[1][i][1]+dist(a[i].s,ed))); 45 } 46 } 47 for(int i=2;i<=n;i++) 48 { 49 for(int j=1;j<=m;j++) 50 { 51 d[i][j][0]=d[i][j][1]=INF; 52 if(a[j].c!=c[i]) continue; 53 for(int k=1;k<=m;k++) 54 { 55 if(a[k].c!=c[i-1]) continue; 56 d[i][j][0]=minn(d[i][j][0],a[j].len+minn(d[i-1][k][0]+dist(a[k].d,a[j].s),d[i-1][k][1]+dist(a[k].s,a[j].s))); 57 d[i][j][1]=minn(d[i][j][1],a[j].len+minn(d[i-1][k][0]+dist(a[k].d,a[j].d),d[i-1][k][1]+dist(a[k].s,a[j].d))); 58 // printf("%d %d d1 = %.3lf d2 = %.3lf ",i,j,d[i][j][0],d[i][j][1]); 59 } 60 if(i==n) mn=minn(mn,minn(d[i][j][0],d[i][j][1]));//dist(a[j].d,ed)dist(a[j].s,ed) 61 } 62 } 63 printf("%.4lf ",mn); 64 } 65 return 0; 66 }
P.S.而且我他人的方法我选的还是很简短的代码,我看oj上大部分人都是2千多Byte的。这2个代码是1千5、6百多Byte。
而我还打了另外一个PG,将颜色隧道按颜色排序了,j和k的循环量比原来的2*N减少了,但是由于排序是O(n log n),N的数据范围也只是几十,所以这个反而速度慢一点。另外。。这个程序只拿了5分,我不知道为何WA,求助啊!●o●
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 #include<cmath> 7 using namespace std; 8 #define M 35 9 #define N 65 10 #define C 105 11 #define INF 2e9 12 13 int m,n; 14 double xs,ys,xt,yt; 15 int q[M],s[C]; 16 struct node{double x,y,xx,yy,l;int c;}; 17 node a[2*N]; 18 double f[2][2*N]; 19 20 void ins(int id,double x,double y,double xx,double yy,int c) 21 { 22 a[id].x=x,a[id].y=y,a[id].c=c; 23 a[id].xx=xx,a[id].yy=yy; 24 a[id].l=sqrt((x-xx)*(x-xx)+(y-yy)*(y-yy)); 25 } 26 bool cmp(node x,node y) {return x.c<y.c;} 27 double mmin(double x,double y) {return x<y?x:y;} 28 double dis(double x,double y,double xx,double yy) {return sqrt((x-xx)*(x-xx)+(y-yy)*(y-yy));} 29 30 int main() 31 { 32 int T,i,j,k; 33 scanf("%d",&T); 34 while (T--) 35 { 36 scanf("%lf%lf%lf%lf",&xs,&ys,&xt,&yt); 37 scanf("%d",&m); 38 for (i=1;i<=m;i++) scanf("%d",&q[i]); 39 scanf("%d",&n); 40 for (i=1;i<=n;i++) 41 { 42 double x,y,xx,yy;int c; 43 scanf("%lf%lf%lf%lf%d",&x,&y,&xx,&yy,&c); 44 ins(2*i-1,x,y,xx,yy,c),ins(2*i,xx,yy,x,y,c); 45 } 46 n*=2; 47 sort(a+1,a+1+n,cmp); 48 49 int c,cc; 50 c=s[0]=a[0].c=0; 51 for (i=1;i<=n;i++) 52 if (a[i].c!=a[i-1].c) 53 { 54 while (c<a[i].c) s[++c]=i-1; 55 s[a[i].c]=i; 56 } 57 s[a[n].c+1]=n+1; 58 59 double ans=INF; 60 q[0]=0,q[m+1]=a[n].c+1; 61 ins(0,xs,ys,xs,ys,0),ins(n+1,xt,yt,xt,yt,a[n].c+1); 62 s[a[n].c+2]=n+2; 63 f[0][0]=0.0; 64 65 int u=1; 66 for (i=1;i<=m+1;i++) 67 { 68 c=q[i],cc=q[i-1]; 69 for (j=s[c];j<s[c+1];j++) 70 { 71 f[u][j]=INF; 72 for (k=s[cc];k<s[cc+1];k++) 73 f[u][j]=mmin(f[u][j],f[1-u][k]+dis(a[k].xx,a[k].yy,a[j].x,a[j].y)+a[j].l); 74 if (i==m+1) ans=mmin(ans,f[u][j]); 75 } 76 u=1-u; 77 } 78 printf("%.4lf ",ans); 79 } 80 return 0; 81 }