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Hdu 1384（差分约束）
题目链接
Intervals
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2931 Accepted Submission(s): 1067Problem DescriptionYou 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
InputThe 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.
OutputThe 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 Input53 7 38 10 36 8 11 3 110 11 1
Sample Output6
令１..i共选ｄ[i]个数，那么有0 <= d[i + 1] - d[i] <= 1;
并且对于每个约束a, b, c， 都有 d[b] - d[a - 1] >= c
这样就是一个线性规划问题，可以用最短路求解。
对于求最大值，就将不等式转化为求最短路中的三角不等式，即：d[u] + w >= d[v]， 然后求最短路即可。
对于求最小值，就将不等式转化为求最长路中的三角不等式，即：d[u] + w <= d[v]， 然后求最长路即可。
当然求最小值时，也可以按求最大值的方法加边，只不过这时候加的边都是反向边，从终点到起点跑最短路，然后结果取反就可以了。
Accepted Code:
1 /************************************************************************* 2 > File Name: 1384.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年08月26日 星期二 08时59分19秒 6 > Propose: 7 ************************************************************************/ 8 #include <queue> 9 #include <cmath>10 #include <string>11 #include <cstdio>12 #include <vector>13 #include <fstream>14 #include <cstring>15 #include <iostream>16 #include <algorithm>17 using namespace std;18 /*Let‘s fight!!!*/19 20 const int INF = 0x3f3f3f3f;21 const int MAX_N = 50050;22 typedef pair<int, int> pii;23 vector<pii> G[MAX_N];24 int n, d[MAX_N];25 bool inq[MAX_N];26 27 void AddEdge(int u, int v, int w) {28 G[u].push_back(pii(v, w));29 }30 31 void spfa(int s) {32 queue<int> Q;33 memset(d, 0x3f, sizeof(d));34 memset(inq, false, sizeof(inq));35 d[s] = 0;36 inq[s] = true;37 Q.push(s);38 while (!Q.empty()) {39 int u = Q.front(); Q.pop(); inq[u] = false;40 for (int i = 0; i < G[u].size(); i++) {41 int v = G[u][i].first, w = G[u][i].second;42 if (d[u] + w < d[v]) {43 d[v] = d[u] + w;44 if (!inq[v]) Q.push(v), inq[v] = true;45 }46 }47 }48 }49 50 int main(void) {51 while (~scanf("%d", &n)) {52 for (int i = 0; i <= 50005; i++) G[i].clear();53 int s = INF, t = -1;54 for (int i = 0; i < n; i++) {55 int a, b, c;56 scanf("%d %d %d", &a, &b, &c);57 b++;58 s = min(s, a); t = max(t, b);59 AddEdge(b, a, -c);60 }61 for (int i = s; i < t; i++) AddEdge(i, i + 1, 1), AddEdge(i + 1, i, 0);62 63 spfa(t);64 65 printf("%d\n", -d[s]);66 }67 68 return 0;69 }
Hdu 1384（差分约束）
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