zoukankan      html  css  js  c++  java
  • poj3680 最大权不相交路径

    Intervals
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 8587   Accepted: 3662

    Description

    You are given N weighted open intervals. The ith interval covers (ai, bi) and weighs wi. Your task is to pick some of the intervals to maximize the total weights under the limit that no point in the real axis is covered more than k times.

    Input

    The first line of input is the number of test case.
    The first line of each test case contains two integers, N and K (1 ≤ KN ≤ 200).
    The next N line each contain three integers ai, bi, wi(1 ≤ ai < bi ≤ 100,000, 1 ≤ wi ≤ 100,000) describing the intervals.
    There is a blank line before each test case.

    Output

    For each test case output the maximum total weights in a separate line.

    Sample Input

    4
    
    3 1
    1 2 2
    2 3 4
    3 4 8
    
    3 1
    1 3 2
    2 3 4
    3 4 8
    
    3 1
    1 100000 100000
    1 2 3
    100 200 300
    
    3 2
    1 100000 100000
    1 150 301
    100 200 300
    

    Sample Output

    14
    12
    100000
    100301

    题意:给你N个区间段的(a,b)和价值,让你在不相交的情况下求m次求得最大值。

    题解:先将每个点都离散化,然后依次添加INF的边,然后输入m条边,用lower_bound找到x,y相应下标,然后加边权值为这个区间段的相反数,跑最小费用流就可以了


    题解这里

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    const int N=500;
    const int M=1e4+88;
    const int INF=0x3f3f3f3f;
    int head[N],tot,pre[N],C[N],F[N],V[N],n,m;
    struct node{
       int u,v,flow,cost,next;
    }e[M];
    void add(int u,int v,int flow,int cost){
       e[tot].u=u;e[tot].v=v;e[tot].flow=flow;e[tot].cost=cost;e[tot].next=head[u];head[u]=tot++;
       e[tot].u=v;e[tot].v=u;e[tot].flow=0;e[tot].cost=-cost;e[tot].next=head[v];head[v]=tot++;
    }
    int SPFA(int s,int t){
       memset(pre,-1,sizeof(pre));
       for(int i=1;i<=t+1;++i) F[i]=0,C[i]=INF,V[i]=0;
       queue<int>Q;
       Q.push(s);
       C[0]=0,F[0]=INF,V[0]=1;
       while(!Q.empty()){
         int u=Q.front();
         Q.pop();
         V[u]=0;
         for(int i=head[u];~i;i=e[i].next){
            int v=e[i].v,f=e[i].flow,c=e[i].cost;
            if(f>0&&C[v]>C[u]+c) {
                C[v]=C[u]+c;
                pre[v]=i;
                F[v]=min(f,F[u]);
                if(!V[v]) V[v]=1,Q.push(v);
            }
         }
       }
       return F[t]&&C[t]!=0;
    }
    int MCMF(int s,int t){
       int ans=0,temp;
       while(temp=SPFA(s,t)){
        for(int i=pre[t];~i;i=pre[e[i].u]) {
            ans+=temp*e[i].cost;
            e[i].flow-=temp;
            e[i^1].flow+=temp;
        }
       }
       return ans;
    }
    struct point{
       int x,y,val;
    }Po[N];
    int ar[N];
    int main(){
       int T;
       for(scanf("%d",&T);T--;){
        scanf("%d%d",&n,&m);
        tot=0;
        memset(head,-1,sizeof(head));
        int ct=0;
        for(int i=1;i<=n;++i) {scanf("%d%d%d",&Po[i].x,&Po[i].y,&Po[i].val);
        ar[++ct]=Po[i].x,ar[++ct]=Po[i].y;
        }
        sort(ar+1,ar+ct+1);
        int num=2;
        for(int i=2;i<=ct;++i) {
            while(ar[i]==ar[num-1]&&i<ct) ++i;
            if(i<=ct) ar[num++]=ar[i];
        }
        for(int i=2;i<=num;++i) add(i-1,i,INF,0);
        add(0,1,m,0);
        add(num,num+1,m,0);
        for(int i=1;i<=n;++i) {
            int l=lower_bound(ar+1,ar+num+1,Po[i].x)-ar;
            int r=lower_bound(ar+1,ar+num+1,Po[i].y)-ar;
            add(l,r,1,-Po[i].val);
        }
        printf("%d
    ",-MCMF(0,num+1));
       }
    }
  • 相关阅读:
    ps插件安装
    CSS3时钟式进度条
    手机web——自适应网页设计(html/css控制)
    7个设计师必备的国际顶尖设计网站
    中​文​字​号​、​磅​和​像​素​对​照​关​系
    图标字体
    用AE如何制作如下三个loading动效,
    u盘装系统
    SpringBoot:Maven创建一个HelloWorld
    eclipse中配置maven和创建第一个 Spring Boot Application
  • 原文地址:https://www.cnblogs.com/mfys/p/7608388.html
Copyright © 2011-2022 走看看