POJ 1201 Intervals

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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

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
const int MAXN = 500000 + 10;
const int MAXM = 500000 * 3 + 10;
int head[MAXN], L, R, n, len;
struct edge{
    int to, next;
    int val;
}e[MAXM];
inline int read(){
    char ch, c;
    int res;
    while (ch = getchar(), ch < '0' || ch > '9') c = ch;
    res = ch - 48;
    while (ch = getchar(), ch >= '0' && ch <= '9')
    res = (res << 3) + (res << 1) + ch - 48;
    if (c == '-') res = -res;
    return res;
}
void add(int from, int to, int val){
    e[len].to = to;
    e[len].val = val;
    e[len].next = head[from];
    head[from] = len++;
}
int dis[MAXN];
int vis[MAXN];
int spfa(){
    queue<int> q;
    for (int i = L; i <= R; i++){
        dis[i] = 0;
        vis[i] = 1;
        q.push(i);
    }
    while (!q.empty()){
        int cur = q.front();
        q.pop();
        vis[cur] = false;
        for (int i = head[cur]; i != -1; i = e[i].next){
            int id = e[i].to;
            if (e[i].val + dis[cur] > dis[id]){
                dis[id] = e[i].val + dis[cur];
                if (!vis[id]){
                    vis[id] = true;
                    q.push(id);
                }
            }
        }
    }
    return dis[R];
}
int main(){
    while (~scanf("%d", &n)){
        memset(head, -1, sizeof(head));
        R = len = 0;
        L = 0x7fffff;
        for (int i = 1; i <= n; i++){
            int from, to, val;
            scanf("%d%d%d", &from, &to, &val);
            R = max(to + 1, R);
            L = min(L, from);
            add(from, to + 1, val);
        }
        for (int i = L; i <= R; i++){
            add(i, i + 1, 0);
            add(i + 1, i, -1);
        }
        printf("%d
", spfa());
    }
    return 0;
}