HDU 3663 Dance chain's non-repeatable coverage

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Test instructions: Give a graph, and then n the power supply range of the city, each now requires each city's D day can have electricity, for city a power, then all the cities adjacent to it will have electricity, but the problem is that each city can only be powered once a day, otherwise it will be broken, And each city has a range of days of power supply and each city can only open the switch once, and can not be used, that is, city A's power supply time must be continuous, there is no need for this city

Idea: Because each city can only be powered once a day, then is not repeated coverage of the dance chain model, then consider what as columns and rows, it is obvious that the problem is n*d+n as a column, N*d is N city D days are to power, and then the next n is to ensure that each city direct power generation, And then consider what as the line, for a city's power generation range, we can intercept all the small areas to be used as rows, of course, not the case to add a 0 0, such as 1 to 5 of the range can become 16 rows, 0 0,1 2,1 3,1 4,1 5,2 2,2 3,2 4,2 5,3 3,3 4, 3 5,4 5,5 5, for each range they even the column is their corresponding number of days of the city, is some small details of the treatment, roughly this is the case, to do not forget the heavy side, heavy edge, heavy side, the main thing said three times

#include <vector> #include <stdio.h> #include <string.h> #include <stdlib.h> #include < Iostream> #include <algorithm>using namespace std;typedef long long ll;typedef unsigned long long ull;const int in f=0x3f3f3f3f;const ll Inf=0x3f3f3f3f3f3f3f3fll;const int Maxn=1010;int l[maxn*400],r[maxn*400],u[maxn*400],d[maxn* 400];//node up and down four-way linked list int c[maxn*400],h[maxn],cnt[maxn],ans[maxn],row[maxn*400];//c column H row cnt list element number int n,m,id,len,d    ; vector<int>g[maxn];void init (int lll) {for (int i=0;i<maxn;i++) g[i].clear (); for (int i=0;i<=lll;i++) {cnt[i]=0;        U[i]=d[i]=i; L[i+1]=i;    r[i]=i+1;    } r[lll]=0;id=lll+1; memset (h,-1,sizeof (H)); void Link (int r,int c) {cnt[c]++; C[id]=c;    Row[id]=r;    U[ID]=U[C];D [U[c]]=id; D[id]=c;    U[c]=id;    if (h[r]==-1) H[r]=l[id]=r[id]=id; else{L[id]=l[h[r]];        R[l[h[r]]]=id; R[ID]=H[R];    L[h[r]]=id; } id++;}    void Remove (int Size) {l[r[size]]=l[size];    R[l[size]]=r[size]; for (int i=d[sIze];i!=size;i=d[i] {for (int j=r[i];j!=i;j=r[j]) {u[d[j]]=u[j];D [u[j]]=d[j];        cnt[c[j]]--; }}}void Resume (int Size) {for (Int. I=d[size];i!=size;i=d[i]) {for (int j=r[i];j!=i;j=r[j]) {U[d[j]]            =j;d[u[j]]=j;        cnt[c[j]]++; }} l[r[size]]=size; R[l[size]]=size;}     int Dance (int k) {int c; if (!         R[0]) {len=k;     return 1;     } for (int min=inf,i=r[0];i;i=r[i]) if (min>cnt[i]) min=cnt[i],c=i;     Remove (c);         for (int i=d[c];i!=c;i=d[i]) {ans[k]=row[i];         for (int j=r[i];j!=i;j=r[j]) Remove (C[j]);         if (Dance (k+1)) return 1;    for (int j=l[i];j!=i;j=l[j]) Resume (C[j]);    } Resume (c); return 0;}    int a[70],b[70],tmp[maxn][2];void linkline (int l,int r,int x,int KKK) {Link (kkk,n*d+x);        for (int i=l;i<=r;i++) {Link (KKK, (i-1) *n+x);            for (unsigned j=0;j<g[x].size (); j + +) {int t=g[x][j];        Link (KKK, (i-1) *n+t);    }}}int Visedge[70][70];int Main () {int u,v;        while (scanf ("%d%d%d", &n,&m,&d)!=-1) {init (n*d+n);        memset (visedge,0,sizeof (Visedge));            for (int i=1;i<=m;i++) {scanf ("%d%d", &u,&v);            if (Visedge[u][v]) continue;            G[u].push_back (v);            G[v].push_back (U);        Visedge[u][v]=1;visedge[v][u]=1;        } for (int i=1;i<=n;i++) scanf ("%d%d", &a[i],&b[i]);        int kkk=1;            for (int i=1;i<=n;i++) {tmp[kkk][0]=0;tmp[kkk++][1]=0;            Link (Kkk-1,n*d+i); for (int j=a[i];j<=b[i];j++) {for (int l=j;l<=b[i];l++) {tmp[kkk][0]=j;tmp[kkk++][1]                    =l;                Linkline (j,l,i,kkk-1);        }}} int flag=dance (0);        if (flag==0) printf ("No solution\n");            else{sort (Ans,ans+len);        for (int i=0;i<len;i++) printf ("%d%d\n", tmp[ans[i]][0],tmp[ans[i]][1]); }        printf ("\ n"); } return 0;}

HDU 3663 Dance chain's non-repeatable coverage