差分约束 poj3159 Candies

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题意：有n个未知数，然后m条信息，每条信息有a，b，c，表示xb-xa<=c，求xn-x1的最大值(n<=3e4)

思路：差分约束，稍微总结了下

B-A<=C 转换成A->B的边权值为C
求B-A最大值转换为求A->B最短路
求B-A最小值转换为求B->A最短路并取负号
如果存在负环，则无解
如果不存在最短路，则无数解

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

const int MX = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;

struct Edge {
    int v, nxt, cost;
} E[MX << 1];
int Head[MX], erear;
int d[MX];
void edge_init(int n) {
    erear = 0;
    for(int i = 1; i <= n; i++) {
        Head[i] = -1;
    }
}
void edge_add(int u, int v, int cost) {
    E[erear].v = v;
    E[erear].cost = cost;
    E[erear].nxt = Head[u];
    Head[u] = erear++;
}
void dijkstra(int u, int n) {
    priority_queue<PII, vector<PII>, greater<PII> >Q;
    for(int i = 1; i <= n; i++) d[i] = INF;
    Q.push(PII(0, u)); d[u] = 0;
    while(!Q.empty()) {
        PII tp = Q.top(); Q.pop();
        int td = tp.first, u = tp.second;
        if(td > d[u]) continue;
        for(int i = Head[u]; ~i; i = E[i].nxt) {
            int v = E[i].v, cost = E[i].cost;
            if(d[u] + cost < d[v]) {
                d[v] = d[u] + cost;
                Q.push(PII(d[v], v));
            }
        }
    }
}

int main() {
    int n, m; //FIN;
    while(~scanf("%d%d", &n, &m)) {
        edge_init(n);
        for(int i = 1; i <= m; i++) {
            int u, v, cost;
            scanf("%d%d%d", &u, &v, &cost);
            edge_add(u, v, cost);
        }
        dijkstra(1, n);
        printf("%d\n", d[n]);
    }
    return 0;
}