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poj 1201 Intervals(差分约束)
2014-06-18 10:59:37      个评论    来源:poj 1201 Intervals(差分约束)  
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Intervals
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 20775 Accepted: 7859

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

题意:告诉你n个区间,要在第i区间至少取ci个数,问所有区间总共至少去多少个数?
这题的构图很是巧妙,把每个边界看成节点,Xi表示0-i有多少个数被选上,则每个区间[li,ri]可以推出X[ri]-X[li-1]>=Ci,即:X[li-1]-X[ri]<=-Ci。
潜在不等式:
1、X[i]-X[i+1]<=0;
2、X[i+1]-X[i]<=0.
本题我多了一个离散化操作。
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 2000100;
struct edge{
    int u , v , d;
    edge(int a = 0 , int b = 0 , int c = 0){
        u = a , v = b , d = c;
    }
}e[maxn];
int head[maxn] , next[maxn] , cnt , n;
int dis[maxn] , vis[maxn] , vt[maxn] , index;
map mp;
struct interval{
    int l , r , value;
    interval(int a = 0 , int b = 0 , int c = 0){
        l = a , r = b , value = c;
    }
};
vector I;

void add(int u , int v , int d){
    e[cnt] = edge(u , v , d);
    next[cnt] = head[u];
    head[u] = cnt++;
}

void initial(){
    I.clear();
    for(int i = 0; i < maxn; i++){
        head[i] = -1;
        dis[i] = INF;
        vis[i] = 0;
        vt[i] = 0;
    }
    cnt = 0;
    index = 0;
    mp.clear();
}

void readcase(){
    int a , b , c;
    for(int i = 0; i < n; i++){
        scanf("%d%d%d" , &a , &b , &c);
        b++;
        I.push_back(interval(a , b , c));
        mp[a] = 0;
        mp[b] = 0;
    }
    map::iterator it;
    for(it = mp.begin(); it != mp.end(); it++){
        it -> second = index++;
    }
    it = mp.begin();
    for(int i = 0; i < index-1; i++){
        int tem = it->first;
        it++;
        add(i+1 , i , 0);
        add(i , i+1 , it->first - tem);
    }
}

bool SPFA(int start){
    queue q;
    q.push(start);
    vt[start]++;
    while(!q.empty()){
        int u = q.front();
        vis[u] = 0;
        if(vt[u]>index) return false;
        q.pop();
        int Next = head[u];
        while(Next != -1){
            int v = e[Next].v , d = e[Next].d;
            if(dis[v] > dis[u]+d){
                dis[v] = dis[u]+d;
                if(vis[v] == 0){
                    vis[v] = 1;
                    vt[v]++;
                    q.push(v);
                }
            }
            Next = next[Next];
        }
    }
    return true;
}

void computing(){
    for(int i = 0; i < I.size(); i++){
        add(mp[I[i].r] , mp[I[i].l] , -1*I[i].value);
    }
    dis[index-1] = 0;
    SPFA(index-1);
    printf("%d\n" , -1*dis[0]);
}

int main(){
    while(scanf("%d" , &n) != EOF){
        initial();
        readcase();
        computing();
    }
    return 0;
}


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