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
);
}
return 0;
}