zoukankan      html  css  js  c++  java
  • POJ3160 Father Christmas flymouse[强连通分量 缩点 DP]

    Father Christmas flymouse
    Time Limit: 1000MS   Memory Limit: 131072K
    Total Submissions: 3241   Accepted: 1099

    Description

    After retirement as contestant from WHU ACM Team, flymouse volunteered to do the odds and ends such as cleaning out the computer lab for training as extension of his contribution to the team. When Christmas came, flymouse played Father Christmas to give gifts to the team members. The team members lived in distinct rooms in different buildings on the campus. To save vigor, flymouse decided to choose only one of those rooms as the place to start his journey and follow directed paths to visit one room after another and give out gifts en passant until he could reach no more unvisited rooms.

    During the days on the team, flymouse left different impressions on his teammates at the time. Some of them, like LiZhiXu, with whom flymouse shared a lot of candies, would surely sing flymouse’s deeds of generosity, while the others, like snoopy, would never let flymouse off for his idleness. flymouse was able to use some kind of comfort index to quantitize whether better or worse he would feel after hearing the words from the gift recipients (positive for better and negative for worse). When arriving at a room, he chould choose to enter and give out a gift and hear the words from the recipient, or bypass the room in silence. He could arrive at a room more than once but never enter it a second time. He wanted to maximize the the sum of comfort indices accumulated along his journey.

    Input

    The input contains several test cases. Each test cases start with two integers N and M not exceeding 30 000 and 150 000 respectively on the first line, meaning that there were N team members living in N distinct rooms and M direct paths. On the next N lines there are N integers, one on each line, the i-th of which gives the comfort index of the words of the team member in the i-th room. Then follow M lines, each containing two integers i and j indicating a directed path from the i-th room to the j-th one. Process to end of file.

    Output

    For each test case, output one line with only the maximized sum of accumulated comfort indices.

    Sample Input

    2 2
    14
    21
    0 1
    1 0

    Sample Output

    35

    Hint

    32-bit signed integer type is capable of doing all arithmetic.

    Source


    题意:多了每个节点分配一个权值,走一条路权值最大

    还是求SCC缩点DP
    很多题解用了spfa求DAG的最长路也可以
     
    //16ms
    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #define C(x) memset(x,0,sizeof(x))
    using namespace std;
    const int N=30005,M=150005;
    inline int read(){
        char c=getchar();int x=0,f=1;
        while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    int n,m,u,v,w[N];
    struct edge{
        int v,ne;
    }e[M];
    int h[N],cnt=0;
    inline void ins(int u,int v){
        cnt++;
        e[cnt].v=v;e[cnt].ne=h[u];h[u]=cnt;
    }
    int dfn[N],low[N],belong[N],dfc,scc,val[N];
    int st[N],top;
    void dfs(int u){
        dfn[u]=low[u]=++dfc;
        st[++top]=u;
        for(int i=h[u];i;i=e[i].ne){
            int v=e[i].v;
            if(!dfn[v]){
                dfs(v);
                low[u]=min(low[u],low[v]);
            }else if(!belong[v])
                low[u]=min(low[u],dfn[v]);
        }
        if(low[u]==dfn[u]){
            scc++;
            while(true){
                int x=st[top--];
                belong[x]=scc;
                if(w[x]>0) val[scc]+=w[x];
                if(x==u) break;
            }
        }
    }
    void SCC(){
        dfc=scc=0;
        C(dfn);C(low);C(belong);C(val);
        top=0;
        for(int i=1;i<=n;i++) if(!dfn[i]) dfs(i);
    }
    
    edge es[N];
    int hs[N],cs=0;
    inline void inss(int u,int v){
        cs++;
        es[cs].v=v;es[cs].ne=hs[u];hs[u]=cs;
    }
    void buildGraph(){
        cs=0;
        C(hs);
        for(int u=1;u<=n;u++){
            int a=belong[u];
            for(int i=h[u];i;i=e[i].ne){
                int b=belong[e[i].v];
                if(a!=b)inss(a,b);
            }
        }
    }
    int f[N];
    int dp(int u){
        if(f[u]!=-1) return f[u];
        f[u]=val[u];
        int mx=0;
        for(int i=hs[u];i;i=es[i].ne){
            int v=es[i].v;//printf("dp %d v %d
    ",u,v);
            mx=max(mx,dp(v));
        }
        return f[u]+=mx;
    }
    int main(int argc, const char * argv[]) {
        while(scanf("%d%d",&n,&m)!=EOF){
            for(int i=1;i<=n;i++) w[i]=read();
            cnt=0;memset(h,0,sizeof(h));
            for(int i=1;i<=m;i++){u=read()+1;v=read()+1;ins(u,v);}
            SCC();
            buildGraph();
            
            memset(f,-1,sizeof(f));
            int ans=0;
            for(int i=1;i<=scc;i++){
                if(f[i]==-1) dp(i);
                ans=max(ans,f[i]);
            }
            printf("%d
    ",ans);
        }
        
        return 0;
    }
  • 相关阅读:
    git忽略.idea文件
    python pip获取所有已安装的第三包
    bootstrap最简单的导航条
    sencha architect开发sencha touch应用注意事项
    反编译sencha toucha打包的apk文件,修改应用名称支持中文以及去除应用标题栏
    TortoiseSVN文件夹及文件图标不显示解决方法
    sql server 约束 查找
    SQLSERVER金额转换成英文大写的函数
    JS把数字金额转换成中文大写数字的函数
    C#中将数字金额转成英文大写金额的函数
  • 原文地址:https://www.cnblogs.com/candy99/p/6026112.html
Copyright © 2011-2022 走看看