【差分约束】SCOI2011糖果

P3275 [SCOI2011]糖果

快noip了我还在干什么啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊

来我们看这道题

根据条件建图， 因为求得是最小值， 所以要跑最长路qwq（这是我记住的QAQ

不想写了让我们直接看看题解吧！

P3275 [SCOI2011]糖果（five20的题解）

有环代表条件冲突， 输出“-1”

建源点的时候反向建会快（这就很妙了

#include<queue>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#define ri register int
#define ll long long
using namespace std;
const int sz = 200010;
int n, k, num = 0;
ll ans;
int head[sz], dis[sz], tot[sz];
bool flag;
bool vis[sz];
struct edge {
    int nxt, to, dis;
}e[sz << 1];
inline int read() {
    char ch = getchar(); int x = 0, f = 1;
    while(ch > '9' || ch < '0') {if(ch == '-') f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0' ; ch = getchar();}
    return x * f;
}
void add(int from, int to, int dis) {
    e[++num].nxt = head[from];
    e[num].to = to;
    e[num].dis = dis;
    head[from] = num;
}
void spfa() {
    queue<int>q;
    q.push(0);
    vis[0] = 1;
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        if(tot[u] == n - 1) {
            printf("-1");
            exit(0);
        }
        vis[u] = 0;
        tot[u]++;
        for(ri i = head[u]; i; i = e[i].nxt) {
            int v = e[i].to;
            if(dis[v] < dis[u] + e[i].dis) {
                dis[v] = dis[u] + e[i].dis;
                if(!vis[v]) {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
}
int main() {
    scanf("%d%d", &n, &k);
    for(ri i = 1; i <= k; i++) {
        int opt = read(), u = read(), v = read();
        if(opt == 1) {
            add(v, u, 0);
            add(u, v, 0);
        }
        else if(opt == 2) {
            if(u == v) {
                flag = 1;
                break;
            }
            else add(u, v, 1);
        }
        else if(opt == 3) add(v, u, 0);
        else if(opt == 4) {
            if(u == v) {
                flag = 1;
                break;
            }
            else add(v, u, 1);
        }
        else if(opt == 5) add(u, v, 0);
     }
     if(flag == 1) {
         printf("-1");
         return 0;
     }
     for(ri i = n; i >= 1; i--) add(0, i, 1);
     spfa();
     for(ri i = 1; i <= n; i++) 
         ans += dis[i];
     printf("%lld", ans);
    return 0;
}
/*
4 7
1 3 2
2 2 4
5 1 3
3 4 2
3 2 3
4 3 1
5 1 4
*/