差分约束系统

变量之间的不等关系转成最短/长路求解。

若A - B <= x

则A <= B + x，此时有一条B->A边权为x的边。

一个形象的图：

不要忘了问题本身的约束条件。

求最小值->最长路

最大值->最短路

负环->无解

不连通->多解

参考资料。

例题：

HDU1384 Intervals 注意有多组数据。

#include <cstdio>
#include <queue>
#include <cstring>

const int N = 100010, INF = 0x3f3f3f3f;

struct Edge {
    int nex, v, len;
    Edge(int N = 0, int V = 0, int L = 0) {
        nex = N;
        v = V;
        len = L;
    }
}edge[2000010]; int tp;

std::queue<int> Q;
int d[N], e[N], a[N], b[N], c[N];
bool vis[N];

inline void add(int x, int y, int z) {
    tp++;
    edge[tp] = Edge(e[x], y, z);
    e[x] = tp;
    return;
}

inline void spfa(int s) {
    memset(d, 0x3f, sizeof(d));
    vis[s] = 1;
    d[s] = 0;
    Q.push(s);
    while(!Q.empty()) {
        int x = Q.front();
        Q.pop();
        vis[x] = 0;
        for(int i = e[x]; i; i = edge[i].nex) {
            int y = edge[i].v;
            if(d[y] > d[x] + edge[i].len) {
                d[y] = d[x] + edge[i].len;
                if(!vis[y]) {
                    vis[y] = 1;
                    Q.push(y);
                }
            }
        }
    }
    return;
}

int main() {
    int n, lm = 0;
    while(scanf("%d", &n) != EOF) {
        memset(e, 0, sizeof(e));
        tp = 0;
        memset(vis, 0, sizeof(vis));
        for(int i = 1; i <= n; i++) {
            scanf("%d%d%d", &a[i], &b[i], &c[i]);
            a[i]++; b[i]++;
            lm = std::max(lm, b[i]);
            /// sum[b[i]] - sum[a[i] - 1] >= c[i]
            add(a[i], b[i] + 1, -c[i]);
        }
        for(int i = 1; i <= lm; i++) {
            add(i, i + 1, 0);
            add(i + 1, i, 1);
        }
        spfa(1);
        printf("%d
", -d[lm + 1]);
    }
    return 0;
}

ZOJ2770 Burn the Linked Camp 一毛一样。多了一个判负环。

#include <cstdio>
#include <queue>
#include <cstring>

typedef long long LL;
const int N = 100010;
const LL INF = 0x3f3f3f3f3f3f3f3fll;

struct Edge {
    int nex, v;
    LL len;
    Edge(int N = 0, int V = 0, LL L = 0) {
        nex = N;
        v = V;
        len = L;
    }
}edge[2000010]; int tp;

std::queue<int> Q;
int e[N], a[N], b[N], c[N], m, n, cnt[N];
LL d[N];
bool vis[N];

inline void add(int x, int y, LL z) {
    tp++;
    edge[tp] = Edge(e[x], y, z);
    e[x] = tp;
    return;
}

inline int spfa(int s) {
    memset(d, 0x3f, sizeof(d));
    memset(cnt, 0, sizeof(cnt));
    memset(vis, 0, sizeof(vis));
    vis[s] = 1;
    d[s] = 0;
    cnt[s] = 1;
    Q.push(s);
    while(!Q.empty()) {
        int x = Q.front();
        Q.pop();
        vis[x] = 0;
        for(int i = e[x]; i; i = edge[i].nex) {
            int y = edge[i].v;
            if(d[y] > d[x] + edge[i].len) {
                d[y] = d[x] + edge[i].len;
                cnt[y] = cnt[x] + 1;
                if(cnt[y] > n + 1) {
                    return -1;
                }
                if(!vis[y]) {
                    vis[y] = 1;
                    Q.push(y);
                }
            }
        }
    }
    return 0;
}

int main() {
    while(scanf("%d%d", &n, &m) != EOF) {
        memset(e, 0, sizeof(e));
        tp = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", &c[i]);
            add(i + 1, i, c[i]);
            add(i, i + 1, 0);
        }
        for(int i = 1; i <= m; i++) {
            scanf("%d%d%d", &a[i], &b[i], &c[i]);
            add(a[i], b[i] + 1, -c[i]);
        }
        int t = spfa(1);
        if(t == -1) {
            printf("Bad Estimations
");
        }
        else {
            printf("%d
", -d[n + 1]);
        }
    }
    return 0;
}

HDU1531 King 注意可能不连通，而每个数也没有非负的限制，所以新建s向每个点连0边保证每个连通块都会被检查一遍。

#include <cstdio>
#include <queue>
#include <cstring>

typedef long long LL;
const int N = 100010;
const LL INF = 0x3f3f3f3f3f3f3f3fll;

struct Edge {
    int nex, v;
    LL len;
    Edge(int N = 0, int V = 0, LL L = 0) {
        nex = N;
        v = V;
        len = L;
    }
}edge[1000010]; int tp;

std::queue<int> Q;
int e[N], a[N], b[N], c[N], m, n, cnt[N];
LL d[N];
bool vis[N];

inline void add(int x, int y, LL z) {
    tp++;
    edge[tp] = Edge(e[x], y, z);
    e[x] = tp;
    return;
}

inline int spfa(int s) {
    memset(d, 0x3f, sizeof(d));
    memset(cnt, 0, sizeof(cnt));
    memset(vis, 0, sizeof(vis));
    vis[s] = 1;
    d[s] = 0;
    cnt[s] = 1;
    Q.push(s);
    while(!Q.empty()) {
        int x = Q.front();
        Q.pop();
        vis[x] = 0;
        for(int i = e[x]; i; i = edge[i].nex) {
            int y = edge[i].v;
            if(d[y] > d[x] + edge[i].len) {
                d[y] = d[x] + edge[i].len;
                cnt[y] = cnt[x] + 1;
                if(cnt[y] > n + 2) {
                    return -1;
                }
                if(!vis[y]) {
                    vis[y] = 1;
                    Q.push(y);
                }
            }
        }
    }
    return 0;
}

char str[3];

int main() {
    //printf("%d 
", (sizeof(edge) + sizeof(cnt) * 7) / 1048576);
    while(scanf("%d", &n), n) {
        scanf("%d", &m);
        memset(e, 0, sizeof(e));
        tp = 0;
        for(int i = 1; i <= m; i++) {
            scanf("%d%d%s%d", &a[i], &b[i], str, &c[i]);
            b[i] += a[i];
            if(str[0] == 'g') {
                c[i]++;
                add(b[i] + 1, a[i], -c[i]);
            }
            else {
                c[i]--;
                add(a[i], b[i] + 1, c[i]);
            }
        }
        for(int i = 1; i <= n + 1; i++) {
            add(n + 2, i, 0);
        }

        int t = spfa(n + 2);
        if(t != -1) {
            printf("lamentable kingdom
");
        }
        else {
            printf("successful conspiracy
");
        }
    }
    return 0;
}

HDU1534 Schedule Problem 之前的都是序列模型，这个直接就是点之间的关系，不需要序列上前缀和。

每个事件的开始时间就是d。

#include <cstdio>
#include <queue>
#include <cstring>

typedef long long LL;
const int N = 100010;
const LL INF = 0x3f3f3f3f3f3f3f3fll;

struct Edge {
    int nex, v;
    LL len;
    Edge(int N = 0, int V = 0, LL L = 0) {
        nex = N;
        v = V;
        len = L;
    }
}edge[1000010]; int tp;

std::queue<int> Q;
int e[N], a[N], b[N], c[N], m, n, cnt[N], Time;
LL d[N];
bool vis[N];

inline void add(int x, int y, LL z) {
    tp++;
    edge[tp] = Edge(e[x], y, z);
    e[x] = tp;
    return;
}

inline int spfa(int s) {
    memset(d, 0x3f, sizeof(d));
    memset(cnt, 0, sizeof(cnt));
    vis[s] = Time;
    d[s] = 0;
    cnt[s] = 1;
    Q.push(s);
    while(!Q.empty()) {
        int x = Q.front();
        Q.pop();
        vis[x] = 0;
        for(int i = e[x]; i; i = edge[i].nex) {
            int y = edge[i].v;
            if(d[y] > d[x] + edge[i].len) {
                d[y] = d[x] + edge[i].len;
                cnt[y] = cnt[x] + 1;
                if(cnt[y] > n + 1) {
                    return -1;
                }
                if(vis[y] != Time) {
                    vis[y] = Time;
                    Q.push(y);
                }
            }
        }
    }
    return 0;
}

char str[4];

int main() {
    //printf("%d 
", (sizeof(edge) + sizeof(cnt) * 7) / 1048576);
    while(scanf("%d", &n), n) {
        memset(e, 0, sizeof(e));
        Time++;
        int x, y;
        tp = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        while(scanf("%s", str)) {
            if(str[0] == '#') break;
            scanf("%d%d", &x, &y);
            if(str[0] == 'F' && str[2] == 'S') {
                add(y, x, a[x]);
            }
            else if(str[0] == 'F') {
                add(y, x, -a[y] + a[x]);
            }
            else if(str[2] == 'F') {
                add(y, x, -a[y]);
            }
            else {
                add(y, x, 0);
            }
        }
        for(int i = 1; i <= n; i++) {
            add(n + 1, i, 0);
        }
        int t = spfa(n + 1);
        printf("Case %d:
", Time);
        if(t == -1) {
            printf("impossible
");
        }
        else {
            for(int i = 1; i <= n; i++) {
                printf("%d %d
", i, -d[i]);
            }
        }
        puts("");
    }
    return 0;
}

HDU1529 Cashier Employment 把每个时间的选用人数做前缀和。发现有7个不等式有三个变量，于是二分其中一个，也就是总共选了多少人。

#include <bits/stdc++.h>

typedef long long LL;
const int N = 110;

struct Edge {
    int nex, v;
    LL len;
    Edge(int N = 0, int V = 0, LL L = 0) {
        nex = N;
        v = V;
        len = L;
    }
}edge[100010]; int tp;

int e[N], cnt[N], stk[N], top, n, m, a[N], vis[N], b[N];
LL d[N];

inline void add(int x, int y, LL z) {
    tp++;
    edge[tp] = Edge(e[x], y, z);
    e[x] = tp;
    return;
}

inline int spfa(int s) {
    static int Time = 0;
    Time++;
    memset(d, 0x3f, sizeof(d));
    memset(cnt, 0, sizeof(cnt));
    stk[++top] = s;
    vis[s] = Time;
    cnt[s] = 1;
    d[s] = 0;
    while(top) {
        int x = stk[top];
        top--;
        vis[x] = 0;
        for(int i = e[x]; i; i = edge[i].nex) {
            int y = edge[i].v;
            if(d[y] > d[x] + edge[i].len) {
                d[y] = d[x] + edge[i].len;
                cnt[y] = cnt[x] + 1;
                if(cnt[y] > n + 1) {
                    return -1;
                }
                if(vis[y] != Time) {
                    vis[y] = Time;
                    stk[++top] = y;
                }
            }
        }
    }
    return 0;
}

inline void clear() {
    memset(b, 0, sizeof(b));
    tp = 0;
    return;
}

inline bool check(int mid) {
    memset(e, 0, sizeof(e));
    tp = 0;
    for(int i = 1; i <= n; i++) {
        add(i, i + 1, 0);
        if(i < 8) {
            add(i + 17, i + 1, -a[i] + mid);
        }
        else {
            add(i - 7, i + 1, -a[i]);
        }
        add(i + 1, i, b[i]);
    }
    add(1, n + 1, -mid);
    add(n + 1, 1, mid);
    int t = spfa(1);
    return t != -1;
}

inline void solve() {
    n = 24;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    scanf("%d", &m);
    for(int i = 1; i <= m; i++) {
        int x;
        scanf("%d", &x);
        b[x + 1]++;
    }
    /// calc
    int l = 0, r = m + 1;
    while(l < r) {
        int mid = (l + r) >> 1;
        if(check(mid)) {
            r = mid;
        }
        else {
            l = mid + 1;
        }
    }
    if(r == m + 1) {
        printf("No Solution
");
        return;
    }
    printf("%d
", r);
    return;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        solve();
        if(T) {
            clear();
        }
    }
    return 0;
}