题意:老板要发年终奖,一共有n个员工,每个人至少888,但还有m个限制条件(a,b),员工a的奖金一定要比员工b多。求老板要满足左右条件最少要发的钱数。
思路:差分约束裸题,只有正权边,为了保证图的连通,初始时将所有点加入队列,并置dis为888,最后求和即可。
另外这个题目用拓扑排序也可以写。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <cstdlib>
#include <set>
#include <string>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 10005;
struct edg{
int v, nxt, w;
}G[maxn << 1];
int tot, pre[maxn];
int n, times[maxn], dis[maxn];
bool vis[maxn];
void add(int u, int v, int w) {
G[tot].v = v;
G[tot].w = w;
G[tot].nxt = pre[u];
pre[u] = tot++;
}
int spfa() {
queue<int> que;
for (int i = 1; i <= n; ++i) {
que.push(i);
vis[i] = true;
dis[i] = 888;
times[i] = 1;
}
while (!que.empty()) {
int u = que.front();
que.pop();
vis[u] = false;
for (int i = pre[u]; ~i; i = G[i].nxt) {
int v = G[i].v, w = G[i].w;
if (dis[u] + w > dis[v]) {
dis[v] = dis[u] + w;
if (!vis[v]) {
if (++times[v] > n) {
return -1;
}
vis[v] = true;
que.push(v);
}
}
}
}
int ans = 0;
for (int i = 1; i <= n; ++i) {
ans += dis[i];
}
return ans;
}
int main(){
int m, a, b;
while (~scanf("%d%d", &n, &m)) {
tot = 0;
memset(pre, -1, sizeof(pre));
while (m--) {
scanf("%d%d", &a, &b);
add(b, a, 1);
}
printf("%d\n", spfa());
}
return 0;
}