POJ 1201 Intervals（差分约束）

POJ 1201

Intervals

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 30971 Accepted: 11990
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：从ai到bi选ci个数，求最小的集合Z
分析：从题目可以读出是差分约束的题，然后推公式。
bi - ai + 1 >= ci
0 <= i + 1 - i <= 1
因为求最小值，所以求一次最长路就行

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int N = 1e6+10;
const int inf = 0x3f3f3f3f;
struct node{
    int u, v, w, nxt;
}edge[N];
int n, cnt, mi, ma;
int fir[N], dis[N], num[N];
bool vis[N];
inline void built(int u, int v, int w){
    edge[cnt] = (node){u, v, w, fir[u]};
    fir[u] = cnt++;
}
inline int spfa(int st){
    memset(vis, false, sizeof(vis));
    vis[st] = true;
    memset(num, 0, sizeof(num));
    num[st]++;
    for(int i = 0; i <= n; i++){
        dis[i] = -inf;
    }
    dis[st] = 0;
    queue<int> Q;
    Q.push(st);
    while(!Q.empty()){
        int u = Q.front();
        Q.pop();
        vis[u] = false;
        for(int i = fir[u]; i; i = edge[i].nxt){
            int v = edge[i].v;
            if(dis[v] < dis[u] + edge[i].w){
                dis[v] = dis[u] + edge[i].w;
                if(!vis[v]){
                    vis[v] = true;
                    Q.push(v);
                    num[v]++;
                    if(num[v] >= n)
                        return -1;
                }
            }
        }
    }
    return dis[ma];
}
int main(){
    #ifdef ONLINE_JUDGE
    #else
        freopen("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    int a, b, c;
    while(~scanf("%d", &n)){
        cnt = 1;
        mi = inf, ma = -inf;
        memset(fir, 0, sizeof(fir));
        for(int i = 1; i <= n; i++){
            scanf("%d%d%d", &a, &b, &c);
            built(a, b + 1, c);
            mi = min(mi, a);
            ma = max(ma, b);
        }
        ma++;
        for(int i = mi; i < ma; i++){
            built(i ,i +  1, 0);
            built(i + 1, i, -1);
        }
        int ans = spfa(mi);
        printf("%d\n", ans);
    }
    return 0;
}