2012杭州网络赛

来源：互联网发布：python split空格分割编辑：程序博客网时间：2024/04/30 09:37

Arrest

费用流。

将每个有敌人的城市拆开，之间连一条容量为1，费用为-inf的边，表示逮捕，保证最后所有敌人被抓到。

源点到起点连一条容量为K，费用为0的边，表示出动K个小分队。

起点到汇点连容量为inf，费用为0的边，表示小分队可以不动，无费用产生。

接着所有城市跑一遍floyd，i<j时连一条edge(i',j,f,w)，i'是拆开的点，容量为inf，费用为distance(i,j)。

起点到各个城市i连一条容量为inf，费用为distance(0,i)的边。

i'到汇点连一条容量为inf，费用为0点，表示结束。

跑一遍费用流就可以了。

/*==================================================*\| 最小费用流 O(V * E * f)| INIT: network g; g.build(v);| CALL: g.mincost(s, t); flow=g.flow; cost=g.cost;| 注意: SPFA增广, 实际复杂度远远小于O(V * E);\*==================================================*/#include <stdio.h>#include <string.h>#define typef int // type of flow#define typec int // type of dis#define min(x,y) (x<y?x:y)const typef inff = 0x000fffff; // max of flowconst typec infc = 0x000fffff; // max of disconst int N=350;const int E=N*N;struct network{    int nv, ne, pnt[E], nxt[E];    int vis[N], que[N], head[N], pv[N], pe[N];    typef flow, cap[E]; typec cost, dis[E], d[N];    void addedge(int u, int v, typef c, typec w) {        pnt[ne] = v; cap[ne] = c;        dis[ne] = +w; nxt[ne] = head[u]; head[u] = (ne++);        pnt[ne] = u; cap[ne] = 0;        dis[ne] = -w; nxt[ne] = head[v]; head[v] = (ne++);    }    int mincost(int src, int sink) {        int i, k, f, r; typef mxf;        for (flow = 0, cost = 0; ; ) {            memset(pv, -1, sizeof(pv));            memset(vis, 0, sizeof(vis));            for (i = 0; i < nv; ++i) d[i] = infc;            d[src] = 0; pv[src] = src; vis[src] = 1;            for (f = 0, r = 1, que[0] = src; r != f; ) {                i = que[f++]; vis[i] = 0;                if (N == f) f = 0;                for (k = head[i]; k != -1; k = nxt[k])                    if(cap[k] && dis[k]+d[i] < d[pnt[k]])                    {                        d[pnt[k]] = dis[k] + d[i];                        if (0 == vis[pnt[k]]) {                        vis[pnt[k]] = 1;                        que[r++] = pnt[k];                        if (N == r) r = 0;                        }                        pv[pnt[k]]=i; pe[pnt[k]]=k;                    }            }            if (-1 == pv[sink]) break;            for (k = sink, mxf = inff; k != src; k = pv[k])                if (cap[pe[k]] < mxf) mxf = cap[pe[k]];            flow += mxf; cost += d[sink] * mxf;            for (k = sink; k != src; k = pv[k]) {                cap[pe[k]] -= mxf; cap[pe[k] ^ 1] += mxf;            }        }        return cost;    }    void build(int v) {        nv = v; ne = 0;        memset(head, -1, sizeof(head));        /*        int x, y; typef f; typec w;        for (int i = 0; i < e; ++i) {            cin >> x >> y >> f >> w; // vertex: 0 ~ n-1            addedge(x, y, f, w);// add arc (u->v, f, w)        }*/    }} g;int f[N][N];int main(){    int n,m,K,i,j,s,t,k;    while ( scanf("%d %d %d",&n,&m,&K),!(n==0&&m==0&&K==0) )    {        //init        g.build(2*n+4);        for (i=0; i<=n; i++)          for (j=0; j<=n; j++)            f[i][j]=inff;        //input        for (i=1; i<=m; i++){            scanf("%d %d %d", &s, &t, &k);            f[s][t]=min(f[s][t],k);            f[t][s]=min(f[t][s],k);        }        //floyd        for (k=0; k<=n; k++)          for (i=0; i<=n; i++)            for (j=0; j<=n; j++)              f[i][j]=min(f[i][j],f[i][k]+f[k][j]);        //build {s=2n+1, t=2n+2}        g.addedge(2*n+1,0,K,0);        for (i=1; i<=n; i++) g.addedge(0,i,inff,f[0][i]);        g.addedge(0,2*n+2,inff,0);        for (i=1; i<=n; i++) {            g.addedge(i,i+n,1,-inff);            g.addedge(i+n,2*n+2,inff,f[i][0]);        }        for (i=1; i<n; i++)          for (j=i+1; j<=n; j++)          {              g.addedge(i+n,j,inff,f[i][j]);          }        g.mincost(2*n+1,2*n+2);        printf("%d\n",g.cost+inff*n);    }    return 0;}

Good Article Good sentence

居然调戏zengxiaoxian。。。

用{A}表示A的所有子串的集合，这题就是求{A}-{B}。

在数学上{A}-{B}={AUB}-{B}={AUB-B}

那么就是求{AUB}中不同子串的个数减去{B}中不同子串的个数。变成模板题了。用SA搞搞，用SAM超时了。T T

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;typedef __int64 ll;#define min(x,y) ( (x)<(y)?(x):(y) )#define max(x,y) ( (x)<(y)?(y):(x) )const int N=311000;//call makesa(); lcp();char *s,sbuff[N];int n, sa[4*N], rank[N], height[N];int buf[4*N], ct[N], sx[N], sax[N];inline bool leq(int a, int b, int x, int y){ return ( (a < x) || (a == x && b <= y) ); }inline bool leq(int a, int b, int c, int x, int y, int z){ return ( (a < x) || (a == x && leq(b, c, y, z))); }inline int geti(int t, int nx, int sa[]){ return ( sa[t]<nx ? sa[t]*3+1 : (sa[t]-nx)*3+2); }static void radix(int a[], int b[], int s[], int n, int k){ // sort a[0..n-1] to b[0..n-1] with keys in 0..k from sint i, t, sum;memset(ct, 0, (k + 1) * sizeof(int));for (i = 0; i < n; ++i) ct[s[a[i]]]++;for (i = 0, sum = 0; i <= k; ++i) {t = ct[i]; ct[i] = sum; sum += t;}for (i = 0; i < n; i++) b[ct[s[a[i]]]++] = a[i];}void suffix(int s[], int sa[], int n, int k){ // !!! require s[n] = s[n+1] = s[n+2] = 0, n >= 2.int i, j, e, p, t;int name = 0, cx = -1, cy = -1, cz = -1;int nx = (n+2)/3, ny = (n+1)/3, nz = n/3, nxz = nx+nz;int *syz = s + n + 3, *sayz = sa + n + 3;for (i=0, j=0; i < n + (nx - ny); i++)if (i%3 != 0) syz[j++] = i;radix(syz , sayz, s+2, nxz, k);radix(sayz, syz , s+1, nxz, k);radix(syz , sayz, s , nxz, k);for (i = 0; i < nxz; i++) {if (s[ sayz[i] ] != cx || s[ sayz[i] + 1 ] != cy ||s[ sayz[i] + 2 ] != cz) {name++; cx = s[ sayz[i] ];cy = s[ sayz[i] + 1 ]; cz = s[ sayz[i] + 2 ];}if (sayz[i] % 3 == 1) syz[ sayz[i] / 3 ] = name;else syz[ sayz[i]/3 + nx ] = name;}if (name < nxz) {suffix(syz, sayz, nxz, name);for (i = 0; i < nxz; i++) syz[sayz[i]] = i + 1;} else {for (i = 0; i < nxz; i++) sayz[syz[i] - 1] = i;}for (i = j = 0; i < nxz; i++)if (sayz[i] < nx) sx[j++] = 3 * sayz[i];radix(sx, sax, s, nx, k);for (p=0, t=nx-ny, e=0; e < n; e++) {i = geti(t, nx, sayz); j = sax[p];if ( sayz[t] < nx ?leq(s[i], syz[sayz[t]+nx], s[j], syz[j/3]) :leq(s[i], s[i+1], syz[sayz[t]-nx+1],s[j], s[j+1], syz[j/3+nx]) ) {sa[e] = i;if (++t == nxz) {for (e++; p < nx; p++, e++)sa[e] = sax[p];}}else {sa[e] = j;if (++p == nx) for (++e; t < nxz; ++t, ++e)sa[e] = geti(t, nx, sayz);}}}void makesa(){memset(buf, 0, 4 * n * sizeof(int));memset(sa, 0, 4 * n * sizeof(int));for (int i=0; i<n; ++i) buf[i] = s[i] & 0xff;suffix(buf, sa, n, 255);
for (int i=0; i<n; i++) rank[ sa[i] ]=i;}
void lcp(){ // O(4 * N)int i, j, k;for (j = rank[height[i=k=0]=0]; i < n - 1; i++, k++)while (k >= 0 && s[i] != s[ sa[j-1] + k ])height[j] = (k--), j = rank[ sa[j] + 1 ];}int len,flg[N];ll getsum(){    int i,t=0,cnt=0,j=0;    ll ret=0;    for (i=0; i<n; i++){        if (s[i]=='$'){            for (; j<=i; j++) flg[j]=i;        }        cnt++;    }    for (i=1; i<n; i++)    {        if ( s[sa[i]]<'a'||s[sa[i]]>'z' ) continue;        t=flg[ sa[i] ]-(height[i]+sa[i]);        if (t>0) {            //printf("i=%d s=%s t=%d\n",i,s+sa[i],t);            ret+=t;        }    }
    //printf("str=%s  substrcnt=%I64d\n",s,ret);    //for (i=0; i<n; i++) printf("%d ",flg[i]);puts("");    return ret;}
//输入一些字符串，用$分开，求出不同的子串个数//call makesa();lcp();getsum();//s=sbuff,n=strlen(s)+1;
int main(){    int m,i,cas=0,len,T,zxx;    ll x,y;    char *sp;    scanf("%d",&T);    while ( T-- )    {        s=sbuff;        memset(flg,0,sizeof(flg));        scanf("%d",&m);        scanf("%s",s);        n=strlen(s)+1;        zxx=n-1;        s[n-1]='$';        s[n]=0;        sp=s+n;        for (i=0; i<m; i++)        {            scanf("%s",sp);            len=strlen(sp);            sp+=len;            *(sp++)='$';        }        *sp=0;        s=sbuff;        n=strlen(s)+1;        makesa();        lcp();        y=getsum();
        s=sbuff+zxx+1;        n=strlen(s)+1;        makesa();        lcp();        x=getsum();
        printf("Case %d: %I64d\n",++cas,(y-x));    }    return 0;}

Finding crosses

模拟题，随便搞搞就好了。

#include <stdio.h>#include <string.h>#define NN 60int n;char s[NN][NN];//图int  vis[NN*NN];int  sum[NN*NN];int  id[NN][NN];//记录连通块idint  hang[NN*NN],lie[NN*NN];void dfs(int i, int j, int c){    if (i<0||j<0||i>=n||j>=n||id[i][j]!=0) return ;    else    if (s[i][j]=='#')    {        id[i][j]=c;        dfs(i+1,j+0,c);        dfs(i+0,j+1,c);        dfs(i-1,j+0,c);        dfs(i+0,j-1,c);    }}int judge(int x, int y, int len){    int i,j,k,sum=0;    i=x,j=y;    //j不动    while (s[i][j]=='#' && i<n) { sum++;i++; }    if (sum!=len) return 0;    i=x+len/2,j=y-len/2,sum=0;    if ( i>=n || j<0 ) return 0;    //i不动    while (s[i][j]=='#' && j<n) { sum++;j++; } //   printf("sum=%d\n",sum);    if (sum!=len) return 0;    return 1;}void init(){        memset(vis,0,sizeof(vis));        memset(sum,0,sizeof(sum));        memset(id,0,sizeof(id));        memset(hang,0,sizeof(hang));        memset(lie,0,sizeof(lie));}int main(){    int T,i,j,k,count,csum;    while ( scanf("%d",&n),n )    {        init();        csum=count=0;        for (i=0; i<n; i++)          scanf("%s",s[i]);        for (i=0; i<n; i++)          for (j=0; j<n; j++)          {              if (s[i][j]=='#' && id[i][j]==0)              {//                  printf("%d %d\n",i,j);                  dfs(i,j,++count);              }          }        for (i=0; i<n; i++)          for (j=0; j<n; j++)            sum[ id[i][j] ]++;        for (i=1;i<=count;i++)            if ( sum[i]>=5 && (sum[i]-1)%4==0 ) vis[i]=1;        for (i=0; i<n; i++)          for (j=0; j<n; j++)            if ( vis[ id[i][j] ] !=0 )            {                csum+=judge(i,j, (sum[ id[i][j] ]+1)/2 );                vis[ id[i][j] ]=0;            }        printf("%d\n",csum);    }    return 0;}

Super Mario

先把输入按照高度给排序了。然后按照高度递增完成insert和query。太久了，大概就是这个思路。开始要查询某区间小于H的个数，转化为按H递增用BIT求区间和。太久了，直接贴代码。

//Binary Indexed Trees\//poj2182//call BIT s;//call s.init(n); n为区间长度。//call s.add(i,v);//call s.sum(i),sum of a[1..i]; call s.sum(x,y),sum of a[x..y];#include <cstdio>#include <string.h>#include <algorithm>using namespace std;#define lowbit(x) ((x) & (-x))typedef int typev;  //type of resstatic const int N=100100;   //max sizestruct BIT{  typev ar[N];      //index 1..N-1  int n;  void init(int nn){    memset(ar,0,sizeof(ar));    n=nn+1;  }  void add(int i, typev v){    for ( ; i<n; ar[i]+=v, i+=lowbit(i) );  }  typev sum(int i){    typev s=0;    for ( ; i>0; s+=ar[i], i-=lowbit(i) );    return s;  }  typev sum(int x, int y){    return (sum(y)-sum(x-1));  }}s;//EOFstruct _node{int x,y,k,id,ans;}a[N];struct _len{int k,x;}l[N];inline bool cmp1(const _node &a, const _node &b){return (a.k<b.k);}inline bool cmp2(const _node &a, const _node &b){return (a.id<b.id);}inline bool cmp3(const _len &a, const _len &b){return (a.k<b.k);}void work(int n, int m){    int i=0,j=0;    while (i<n && j<m)    {        while (i<n&&l[i].k<=a[j].k) {            s.add(l[i++].x,1);        }        while (j<m&&i<n&&l[i].k>a[j].k){            a[j].ans=s.sum(a[j].x,a[j].y);            j++;        }    }    while (j<m){        a[j].ans=s.sum(a[j].x,a[j].y);        j++;    }}int main(){    int n,m,i,cas=0,T;    scanf("%d",&T);    while (T--){        scanf("%d %d", &n, &m);        s.init(n+10);        for (i=0; i<n; i++){            scanf("%d",&l[i].k);            l[i].x=i+1;        }        for (i=0; i<m; i++) {            scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].k);            a[i].x++,a[i].y++;            a[i].id=i;        }        sort(l,l+n,cmp3);        sort(a,a+m,cmp1);        work(n,m);        sort(a,a+m,cmp2);        printf("Case %d:\n",++cas);        for (i=0; i<m; i++)        {            printf("%d\n",a[i].ans);        }    }    return 0;}