差分约束

1.求最小值的情况。

将关系都化成 a[ i ] - a[ j ] >= k

然后 j 向 i 建立一个价值k的有向边。

然后跑一个最长路。

例如：POJ-1201

#include
#include
#include
#include
#include
#include
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 5e4 + 10;
int head[N], to[N<<2], ct[N<<2], nt[N<<2];
int tot;
void add(int u, int v, int w){
    to[tot] = v;
    nt[tot] = head[u];
    ct[tot] = w;
    head[u] = tot++;
}
int d[N+100];
int vis[N+100];
queue<int> q;
void spfa(){
    vis[0] = 1;
    q.push(0);
    while(!q.empty()){
        int now = q.front();
        q.pop();
        vis[now] = 0;
        for(int i = head[now];~i; i = nt[i]){
            int v = to[i];
            if(d[v] < d[now] + ct[i]){
                d[v] = d[now] + ct[i];
                if(!vis[v]){
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }
    }
}
int main(){
    memset(head, -1, sizeof head);
    int n, a, b, c;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i){
        scanf("%d%d%d", &a, &b, &c);
        add(a, b+1, c);
    }
    for(int i = 0; i + 1 < N; ++i){
        add(i, i+1, 0);
        add(i+1, i, -1);
    }
    memset(d, _inf, sizeof d);
    d[0] = 0;
    spfa();
    printf("%d\n", d[N-1]);
    return 0;
}

View Code

 

2.求最大值的情况

将关系都转换成a[i]-a[j]<=k

然后j向i建立一个价值k的有向边。

跑一个最短路。

例如：HDU-3440

#include
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 2e5 + 100;
int head[1010];
int to[N], ct[N], nt[N];
int tot;
void add(int u, int v, int w){
    to[tot] = v; ct[tot] = w;
    nt[tot] = head[u]; head[u] = tot++;
}
void init(){
    memset(head, -1, sizeof head);
    tot = 0;
}
pll p[N];
int d[1010], cnt[1010], vis[1010], n;
int spfa(int s, int t){
    memset(d, inf, sizeof d);
    memset(cnt, 0, sizeof cnt);
    memset(vis, 0, sizeof vis);
    d[s] = 0;
    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = 0;
//        cout << u << endl;
        for(int i = head[u]; ~i; i = nt[i]){
            int v = to[i];
            if(d[v] > d[u] + ct[i]){
                d[v] = d[u] + ct[i];
                if(!vis[v]){
                    vis[v] = 1;
                    ++cnt[v];
                    if(cnt[v] > n) return -1;
                    q.push(v);
                }
            }
        }
    }
    return d[t];
}
int main(){
    int d;
    int T;
    scanf("%d", &T);
    for(int _cas = 1; _cas <= T; ++_cas){
        init();
        /// x[i] - x[i-1] >= 1
        /// x[i-1] - x[i] <= -1
        scanf("%d%d", &n, &d);
        for(int i = 1; i < n; ++i)
            add(i+1, i, -1);
        for(int i = 1; i <= n; ++i){
            scanf("%d", &p[i].fi);
            p[i].se = i;
        }
        sort(p+1, p+1+n);
        for(int i = 1; i < n; ++i){
            int u = min(p[i].se, p[i+1].se);
            int v = u ^ p[i].se ^ p[i+1].se;
            /// x[v] - x[u] <= d
            add(u, v, d);
        }
        int s = min(p[1].se, p[n].se);
        int t = s ^ p[1].se ^ p[n].se;
        printf("Case %d: %d\n", _cas,spfa(s,t));
    }
    return 0;
}

View Code

 

关键就是利用了三角形不等式。 因此可以用spfa来算最短/长路来解决。

需要注意的就是 可能会有一个负环/正环来无限修改答案，记得记录次数。

转载于:https://www.cnblogs.com/MingSD/p/11132744.html