HOJ 13415 Website Tour(强连通缩点+多重背包dp)

来源：互联网发布：淘宝商城童装毛衣女孩编辑：程序博客网时间：2024/06/03 23:07

题意：每个网站有个广告，看这个广告要花t的时间，能得p的分，最多只能看s次，然后给一些网站以及它所能到达的网站，从一个网站到另一个是不花时间的，但是不能停在同一个网站连续不停的看。现在有T的时间，起点任意，问最大得分。

做法：由于都是有向边，所以我们可以把强连通分量缩点了，这其中的点都可以互相到达，再把这些缩后的点给拓扑排序，就是个DAG了，我们就可以dp了，在这个连通分量的点里面就是多重背包了，可以用二进制分解的办法做到每个点Tlogk的复杂度。然后再去更新它之后的点即可。总复杂度o(n*T*logk+m)。

AC代码：

#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<ctype.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdlib>#include<stack>#include<queue>#include<set>#include<map>#include<cmath>#include<ctime>#include<string.h>#include<string>#include<sstream>#include<bitset>using namespace std;#define ll __int64#define ull unsigned __int64#define eps 1e-8#define NMAX 1000000000#define MOD (1<<30)#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define PI acos(-1)template<class T>inline void scan_d(T &ret){    char c;    int flag = 0;    ret=0;    while(((c=getchar())<'0'||c>'9')&&c!='-');    if(c == '-')    {        flag = 1;        c = getchar();    }    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();    if(flag) ret = -ret;}const int maxn = 100+10;const int maxm = 1000+10;struct Edge{    int v,next;}e[2][maxm];int head[2][maxn], ecnt[2];void add_edge(int u, int v, int d){    e[d][ecnt[d]].v = v; e[d][ecnt[d]].next = head[d][u];    head[d][u] = ecnt[d]++;}struct node{    int p,t,k;    node(){}    node(int _p, int _t, int _k):p(_p),t(_t),k(_k){}}no[maxn];bool zi[maxn];int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt, tp;int S[maxn];void dfs(int u){    pre[u] = lowlink[u] = ++dfs_clock;    S[++tp] = u;    for(int i = head[0][u]; ~i; i = e[0][i].next)    {        int v = e[0][i].v;        if(!pre[v])        {            dfs(v);            lowlink[u] = min(lowlink[u],lowlink[v]);        }        else if(!sccno[v]) lowlink[u] = min(lowlink[u],pre[v]);    }    if(lowlink[u] == pre[u])    {        scc_cnt++;        while(1)        {            int x = S[tp]; tp--;            sccno[x] = scc_cnt;            if(x == u) break;        }    }}void find_scc(int n){    tp = -1;    dfs_clock = scc_cnt = 0;    memset(sccno,0,sizeof(sccno));    memset(pre,0,sizeof(pre));    for(int i = 1; i <= n; i++) if(!pre[i])        dfs(i);}vector<int>vec[105];int Em[maxm][2];bool bian[maxn][maxn];int in[maxn],top[maxn];void topo(int n){    queue<int>q;    for(int i = 1; i <= n; i++) if(in[i] == 0)        q.push(i);    int k = 1;    while(!q.empty())    {        int x = q.front(); q.pop();        top[k++] = x;        for(int i = head[1][x]; ~i ; i = e[1][i].next)        {            int v = e[1][i].v;            in[v]--;            if(in[v] == 0) q.push(v);        }    }}int dp[maxn][10000+10];node ha[maxn];int main(){#ifdef GLQ    freopen("input.txt","r",stdin);//    freopen("o.txt","w",stdout);#endif    int n,m,T;    while(~scanf("%d%d%d",&n,&m,&T) && n+m+T)    {        memset(head,-1,sizeof(head));        memset(ecnt,0,sizeof(ecnt));        memset(zi,0,sizeof(zi));        for(int i = 1; i <= n; i++)            scanf("%d%d%d",&no[i].p,&no[i].t,&no[i].k);        for(int i = 1; i <= m; i++)        {            int u,v;            scanf("%d%d",&u,&v);            if(u == v)            {                zi[u] = 1;                Em[i][0] = -1;            }            else            {                Em[i][0] = u;                Em[i][1] = v;                add_edge(u,v,0);            }        }        find_scc(n);        for(int i = 1; i <= scc_cnt; i++) vec[i].clear();        for(int i = 1; i <= n; i++)        {            vec[sccno[i]].push_back(i);        }        for(int i = 1; i <= scc_cnt; i++) if(vec[i].size() == 1 && zi[vec[i][0]] == 0)            no[vec[i][0]].k = 1;        memset(bian,0,sizeof(bian));        memset(in,0,sizeof(in));        for(int i = 1; i <= m; i++) if(Em[i][0] != -1)        {            int u = sccno[Em[i][0]], v = sccno[Em[i][1]];//            cout<<u<<" "<<v<<endl;            if(u != v && bian[u][v] == 0)            {                bian[u][v] = 1;                in[v]++;                add_edge(u,v,1);            }        }        topo(scc_cnt);        for(int i = 1; i <= scc_cnt; i++)            for(int j = 0; j <= T; j++)                dp[i][j] = 0;        for(int i = 1; i <= scc_cnt; i++)        {            int pos = top[i];            int sz = vec[pos].size();            for(int j = 0; j < sz; j++)            {                int id = vec[pos][j];                if(no[id].k*no[id].t >= T)                {                    for(int v = no[id].t; v <= T; v++)                        dp[pos][v] = max(dp[pos][v],dp[pos][v-no[id].t]+no[id].p);                    continue;                }                int cnt = 0;                while((1<<cnt+1)-1 < no[id].k)                {                    ha[cnt] = node((1<<cnt)*no[id].p, (1<<cnt)*no[id].t, (1<<cnt));                    cnt++;                }                int tt = no[id].k-(1<<cnt)+1;                ha[cnt++] = node(tt*no[id].p, tt*no[id].t, tt);                for(int k = 0; k < cnt; k++)                {                    for(int v = T; v >= ha[k].t; v--)                        dp[pos][v] = max(dp[pos][v],dp[pos][v-ha[k].t]+ha[k].p);                }            }            for(int j = head[1][pos]; ~j; j = e[1][j].next)            {                int v = e[1][j].v;                for(int val = 0; val <= T; val++)                    dp[v][val] = max(dp[v][val],dp[pos][val]);            }        }        int ans = 0;        for(int i = 1; i <= scc_cnt; i++)            ans = max(ans,dp[i][T]);        printf("%d\n",ans);    }    return 0;}

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