zoj 1508 差分约束 Bellman-Ford

Intervals

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.

Write a program that:

> reads the number of intervals, their endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi - ai + 1.

Process to the end of file.

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

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Source: Southwestern Europe 2002

参考了《数与图的完美结合--浅析差分约束系统》，by 张威：http://pan.baidu.com/share/link?shareid=508660&uk=2720516383&fid=743287535

对属于区间[ai, bi]的任意元素,按属于于不属于题述序列关系，可定义如下映射：

t=0,表某元素不在序列中；1,表在序列中。

则可令Si= 从ai-1到bi,求和（t）,即Si表该区间[ai,bi]中在序列的元素个数。至此，不难构造不等式（组）：

S(bi)-S(ai-1)>=ci；另外由任意S(i)与S(i-1)的关系也可得：

0<=S(i)-S(i-1)<=1; 转化为：

S(ai-1) - S(bi) <= -ci

S(i-1) - S(i) <= 0

S(i) - S(i-1) <= 1

于是就可以直接建图用bellmanford算法了，但答案是多少呢？这里我不很清楚，可能是这样建图产生负权边，然后求相对值吧。论文给出的是d[最大节点标号]-d[最小节点标号，即0].

AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<queue>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))

#define MAXN 66666
#define MAXM 200000
#define INF 100000

struct edge{
    int u,v,nxt,w;
}e[MAXM];
int cc,h[MAXN];
int inq[MAXN],d[MAXN];
int ans;

inline void add(int u, int v, int w){
    e[cc]=(edge){u,v,h[u],w};
    h[u]=cc++;
}

queue<int> q;
int bford(int s, int n){
    int u,i,w,v;
    while(!q.empty()) q.pop();  q.push(s);
 //   memset(cnt, 0, sizeof(cnt));    cnt[s]=1;
    memset(inq, 0, sizeof(inq));    inq[s]=1;
    for(i=0; i<n; i++) d[i]=INF;    d[s]=0;
    while(!q.empty()){
        u=q.front();    q.pop();    inq[u]=0;
        for(i=h[u]; i!=-1; i=e[i].nxt){
            v=e[i].v;
            w=e[i].w;
            if(d[v]>d[u]+w){
                d[v]=d[u]+w;
                if(!inq[v]){
    //                if(++cnt[v]>n) return 1;
                    q.push(v);
                    inq[v]=1;
                }
            }
        }
    }
    ans=d[s-1]-d[0];            // 结果不是d[s-1]么？。。。
    return 0;
}

#define POINT(x) ((x) < 0 ? 0 : (x) )

int main(){
//  freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    int n;
    while(scanf(" %d",&n)==1){
        int a,b,c;
        int i,ub=0;
        cc=0;
        memset(h, -1, sizeof(h));
        for(i=0; i<n; i++){
            scanf(" %d %d %d",&a,&b,&c);
            add(b, POINT(a-1), -c);         // S(bi) - S(ai-1) <= C(i) --如果a-1<0,则用0代替a-1，仍满足,此时0表示ai-1项不存在
            ub=MAX(ub, b);                  // 记录最大点编号
        }
        ub++;
        for(i=1; i<ub; i++){
            add(i, i-1, 0);                 // S(i-1) - S(i) <= 0
            add(i-1, i, 1);                 // S(i) - S(i-1) <= 1
        }
        for(i=0; i<ub; i++) add(ub, i, 0);  // S(ub+1) - i <= 0  --虚拟源点，编号ub+1

        bford(ub, ub+2);                    // 共有ub+2个点
        printf("%d\n", ans);
    }
    return 0;
}