POJ-1716 Integer Intervals (differential constraint system)

POJ-1716 Integer Intervals (differential constraint system)

Given N intervals, you need to find the number of M, which satisfies the fact that there are at least two different numbers in each interval.

Solution: I still do not know much about the difference constraint system, and I am not very good at mathematics.
Drawing on others' ideas, I feel a little bit DP thinking
Set d [I] to [0, I-1]. The number of d [I] In this interval meets the requirements.
Given a range [a, B], d [B + 1]-d [a]> = 2 (B + 1 because B is also included in the range)
Convert it to d [a]-d [B + 1] <=-2, which is the first sub-Statement
Then, in the interval [I, I + 1 ],
D [I + 1]-d [I]> = 0 to d [I]-d [I + 1] <= 0
D [I + 1]-d [I] <= 1
Three sub-charts
Use Max (rightmost boundary + 1) as the source point for SPFA
The final answer is d [Max]-d [Min].

#include 
  
   #include 
   
    #include 
    
     using namespace std;    #define N 10010#define M 30010#define INF 0x3f3f3f3fstruct Edge{    int dist, to, next;}E[M];int head[N], d[N], n, tot;bool vis[N];void AddEdge(int u, int v, int dist) {    E[tot].to = v;    E[tot].dist = dist;    E[tot].next = head[u];    head[u] = tot++;}void SPFA(int s) {    for (int i = 0; i <= s; i++) {        d[i] = INF;        vis[i] = 0;    }    queue
     
       q;    q.push(s);    d[s] = 0;    while (!q.empty()) {        int u = q.front();        q.pop();        vis[u] = false;        for (int i = head[u]; i != -1; i = E[i].next) {            int v = E[i].to;            if (d[v] > d[u] + E[i].dist) {                d[v] = d[u] + E[i].dist;                if (!vis[v]) {                    vis[v] = true;                    q.push(v);                }            }        }    }}void solve() {    memset(head, -1, sizeof(head));    tot = 0;    int Min = INF, Max = -INF;    int u, v;    for (int i = 0; i < n; i++) {        scanf(%d%d, &u, &v);        Min = min(Min, u);        Max = max(Max, v + 1);        AddEdge(v + 1, u, -2);    }    for (int i = 0; i < Max; i++) {        AddEdge(i, i + 1, 1);        AddEdge(i + 1, i, 0);    }    SPFA(Max);    printf(%d, d[Max] - d[Min]);}int main() {    while (scanf(%d, &n) != EOF) {        solve();    }    return 0;}