BZOJ 2330

传送门

题目分析

差分约束 这里做个简单介绍：形如(x_i - x_j >= d)的不等式，可以联想到我们求最短路时(d_v <= d_u + len)，则上式可以变形为(x_i >= x_j + d)即连一条j->i的长度为d的边并跑最长路，dis[i]则是满足条件的最小解（因为上面等式采用的>=号，所以求出的时最小解，同理当变形为(x_j <= x_i - d) 采用<= 时求出的是最大解）。

转差分约束

这道题也是经典的差分约束，只是要注意几个问题：

• spfa判负环 无解
• 输入矛盾条件时直接无解

code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
 
const int N = 1e6 + 5;
typedef long long ll;
const ll OO = 0x3f3f3f3f;
int times[N];
int n, k;
 
ll dis[N];
int ecnt, adj[N], go[N << 2], nxt[N << 2], len[N << 2];
bool vst[N];
 
inline void addEdge(int u, int v, int l){
    nxt[++ecnt] = adj[u], adj[u] = ecnt, go[ecnt] = v, len[ecnt] = l;
}
 
inline int read(){
    int i = 0, f = 1;char ch = getchar();
    for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
    if(ch == '-') ch = getchar(), f = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        i = (i << 3) + (i << 1) + (ch - '0');
    return i * f;
}
 
inline void wr(ll x){
    if(x < 0) putchar('-'), x = -x;
    if(x > 9) wr(x / 10);
    putchar(x % 10 + '0');
}
 
inline bool spfa(){
    static int que[N], qn;
    for(int i = 1; i <= n; i++) dis[i] = -OO;
    dis[0] = 0;
    que[qn = 1] = 0;
    vst[0] = true;
    for(int ql = 1; ql <= qn; ql++){
        int u = que[ql];
        vst[u] = false; 
        times[u]++;
        if(times[u] == n) return false;
        for(int e = adj[u]; e; e = nxt[e]){
            int v = go[e];
            if(dis[v] < dis[u] + len[e]){
                dis[v] = dis[u] + len[e];
                if(!vst[v]) vst[v] = true, que[++qn] = v;
            }
        }
    }
    return true;
}
 
int main(){
    n = read(), k = read();
    for(int i = n; i >= 1; i--) addEdge(0, i, 1);
    for(int i = 1; i <= k; i++){
        int x = read(), a = read(), b = read();
        switch(x){
            case 1:{
                if(a != b){
                    addEdge(a, b, 0); 
                    addEdge(b, a, 0); 
                }
                break;
            }
            case 2:{
                if(a == b){
                    printf("-1");
                    return 0;
                }
                addEdge(a, b, 1); 
                break;
            }
            case 3:{
                if(a != b)
                    addEdge(b, a, 0); 
                break;
            }
            case 4:{
                if(a == b){
                    printf("-1");
                    return 0;
                }
                addEdge(b, a, 1); 
                break;
            }
            case 5:{
                if(a != b)
                    addEdge(a, b, 0); 
                break;
            }
            default: break;
        }
    }
    if(!spfa()){
        printf("-1");
        return 0;
    }
    ll ans = 0;
    for(int i = 1; i <= n; i++){
        ans += dis[i];
    }
    wr(ans);
    return 0;
}