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题目链接：http://poj.org/problem?id=1201
题目大意：
有一个序列，题目用n个整数组合 [ai，bi，ci]来描述它，[ai，bi，ci]表示在该序列中处于[ai，bi]这个区间的整数至少有ci个。如果存在这样的序列，请求出满足题目要求的最短的序列长度是多少。
思路：
此题为差分约束
做题时遵循三个不等式
1.S[b] - S[a - 1] >= c
2.S[b] - s[b - 1] >= 0
3.S[b - 1] - S[b] >= -1
然后建图就ok了owo
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <queue> 5 #include <cstring> 6 using namespace std; 7 const int MAXN = 500000 + 10; 8 const int MAXM = 500000 * 3 + 10; 9 int head[MAXN], L, R, n, len; 10 struct edge{ 11 int to, next; 12 int val; 13 }e[MAXM]; 14 inline int read(){ 15 char ch, c; 16 int res; 17 while (ch = getchar(), ch < '0' || ch > '9') c = ch; 18 res = ch - 48; 19 while (ch = getchar(), ch >= '0' && ch <= '9') 20 res = (res << 3) + (res << 1) + ch - 48; 21 if (c == '-') res = -res; 22 return res; 23 } 24 void add(int from, int to, int val){ 25 e[len].to = to; 26 e[len].val = val; 27 e[len].next = head[from]; 28 head[from] = len++; 29 } 30 int dis[MAXN]; 31 int vis[MAXN]; 32 int spfa(){ 33 queue<int> q; 34 for (int i = L; i <= R; i++){ 35 dis[i] = 0; 36 vis[i] = 1; 37 q.push(i); 38 } 39 while (!q.empty()){ 40 int cur = q.front(); 41 q.pop(); 42 vis[cur] = false; 43 for (int i = head[cur]; i != -1; i = e[i].next){ 44 int id = e[i].to; 45 if (e[i].val + dis[cur] > dis[id]){ 46 dis[id] = e[i].val + dis[cur]; 47 if (!vis[id]){ 48 vis[id] = true; 49 q.push(id); 50 } 51 } 52 } 53 } 54 return dis[R]; 55 } 56 int main(){ 57 while (~scanf("%d", &n)){ 58 memset(head, -1, sizeof(head)); 59 R = len = 0; 60 L = 0x7fffff; 61 for (int i = 1; i <= n; i++){ 62 int from, to, val; 63 scanf("%d%d%d", &from, &to, &val); 64 R = max(to + 1, R); 65 L = min(L, from); 66 add(from, to + 1, val); 67 } 68 for (int i = L; i <= R; i++){ 69 add(i, i + 1, 0); 70 add(i + 1, i, -1); 71 } 72 printf("%d ", spfa()); 73 } 74 return 0; 75 }