POJ -- 1201--Intervals (差分约束系统)

 Intervals

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1201

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

53 7 38 10 36 8 11 3 110 11 1

Sample Output

6

题意：要求在给出的 n 个区间中，每个区间 [a, b] 至少取 c 个数，问满足要求的集合中元素个数

           最少是多少？

分析：dis[x+1] => 0--x 中有 dis[x+1] 个满足条件的

            2 个式子：dis[b+1] - dis[a] >= c

            0 <= dis[u+1] - dis[u] <= 1 => dis[u+1] - dis[u] >= 0  and dis[u] - dis[u+1] >= -1.

代码如下：

1：用 vector 建边 （时间太长了 1766 ms）

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;#define ll unsigned long long #define inf 0x3f3f3f3f#define N 50005int vis[N],dis[N];struct r{int v, w;}tmp;vector<r> p[N];void spfa(int mi){memset(vis,0,sizeof(vis));memset(dis,-inf,sizeof(dis));queue<int> q;q.push(mi);vis[mi]=1;dis[mi]=0;while(!q.empty()){int out=q.front(); q.pop();vis[out]=0;for(int i=0;i<p[out].size();i++){if(dis[p[out][i].v]<dis[out]+p[out][i].w){//求最长路dis[p[out][i].v]=dis[out]+p[out][i].w;if(!vis[p[out][i].v]){q.push(p[out][i].v);vis[p[out][i].v]=1;}}}}}int main(){#ifdef yibinfreopen("in.txt","r",stdin);#endifint t, n, m, i, j, k, a, b, c;while(~scanf("%d", &t)){int mi=inf, ma=-inf;while(t--){scanf("%d%d%d",&a, &b, &c);mi=min(mi, a);ma=max(ma, b+1);tmp.v=b+1, tmp.w=c;//dis[b+1]-dis[a]>=cp[a].push_back(tmp);}//0<=dis[u+1]-dis[u]<=1for(i=mi;i<=ma;i++){tmp.v=i+1, tmp.w=0;p[i].push_back(tmp);//dis[u+1]-dis[u]>=0tmp.v=i, tmp.w=-1;p[i+1].push_back(tmp);//dis[u]-dis[u+1]>=-1}spfa(mi);printf("%d\n",dis[ma]);}return 0;}

2：改用前向星建边 （时间明显减少了 共 344ms（C++）250ms ( G++ )）

   （就是数组要开至少 3 倍，小了一直 RE， 不太懂！）

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;#define inf 0x3f3f3f3f#define N 50005*3//！数组注意开至少 3 倍！int vis[N], dis[N], head[N], k;struct r{int v, w, next;}p[N];void add(int u, int v, int w)//前向星建边{p[k].v = v;p[k].w = w;p[k].next = head[u];head[u] = k++;}void spfa(int mi){memset(vis, 0, sizeof(vis));memset(dis, -inf, sizeof(dis));queue<int> q;q.push(mi);vis[mi] = 1;dis[mi] = 0;while (!q.empty()){int out = q.front(); q.pop();vis[out] = 0;for (int i = head[out]; i + 1; i = p[i].next){if (dis[p[i].v] < dis[out] + p[i].w){//求最长路dis[p[i].v] = dis[out] + p[i].w;if (!vis[p[i].v]){q.push(p[i].v);vis[p[i].v] = 1;}}}}}int main(){#ifdef OFFLINEfreopen("t.txt", "r", stdin);#endifint t, n, m, i, j, a, b, c;while (~scanf("%d", &t)){int mi = inf, ma = -inf, k = 0;memset(head, -1, sizeof(head));while (t--){scanf("%d%d%d", &a, &b, &c);mi = min(mi, a);ma = max(ma, b + 1);add(a, b + 1, c);//dis[b+1]-dis[a]>=c}//0<=dis[u+1]-dis[u]<=1for (i = mi; i <= ma; i++){add(i, i + 1, 0);//dis[u+1]-dis[u]>=0add(i + 1, i, -1);//dis[u]-dis[u+1]>=-1}spfa(mi);printf("%d\n", dis[ma]);}return 0;}