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  • POJ 1201 Intervals

    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 28852   Accepted: 11131

    Description

    You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn. 
    Write a program that: 
    reads the number of intervals, their end points and integers c1, ..., cn from the standard input, 
    computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,...,n, 
    writes the answer to the standard output. 

    Input

    The first line of the input contains an integer n (1 <= n <= 50000) -- the number of intervals. The following n lines describe the intervals. The (i+1)-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.

    Output

    The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i=1,2,...,n.

    Sample Input

    5
    3 7 3
    8 10 3
    6 8 1
    1 3 1
    10 11 1

    Sample Output

    6

    Source

    #include <map>
    #include <set>
    #include <stack>
    #include<queue>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define eps 1e-15
    #define MAXN  50005
    #define INF 1000000007
    #define MAX(a,b)(a>b?a:b)
    #define MIN(a,b)(a<b?a:b)
    int N,n,Max,Min;
    int head[MAXN],d[MAXN],vis[MAXN];
    struct EDG{ int v,w,next; }edge[3*MAXN];
    void AddEdge(int u,int v,int w){
        edge[N].v=v;edge[N].w=w;edge[N].next=head[u];head[u]=N++;
    }
    void SPFA(){
        for(int i=Min;i<=Max;i++) d[i]=-INF;
        d[Min]=0;queue<int>q;q.push(Min);
        while(!q.empty()){
            int x=q.front();q.pop();
            vis[x]=0;
            for(int e=head[x];e!=-1;e=edge[e].next)
                if(d[edge[e].v]<d[x]+edge[e].w){
                    d[edge[e].v]=d[x]+edge[e].w;
                    if(!vis[edge[e].v]){
                        q.push(edge[e].v);
                        vis[edge[e].v]=1;
                    }
                }
        }
    } 
    int main(){
        while(scanf("%d", &n)!=EOF){
            memset(vis,0,sizeof(vis));
            memset(edge,0,sizeof(edge));
            memset(head,-1,sizeof(head));
            Min=INF;Max=-INF;N=0;
            for(int i=0;i<n;i++){
                int u,v,w;
                scanf("%d%d%d",&u,&v,&w);
                AddEdge(u,v+1,w);
                Min=MIN(Min,u);
                Max=MAX(Max,v+1);
            }
            for(int i=Min;i<Max;i++){
                AddEdge(i,i+1,0);
                AddEdge(i+1,i,-1);
            }
            SPFA();
            printf("%d
    ", d[Max]);
        }
    } 
    /*
    5
    3 7 3
    8 10 3
    6 8 1
    1 3 1
    10 11 1
    */
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  • 原文地址:https://www.cnblogs.com/cangT-Tlan/p/9092107.html
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