[NOI2010]海拔——最小割+对偶图

题目链接

SOLUTION

想一下最优情况下肯定让平路或下坡尽量多，于是不难想到这样构图：包括左上角的一部分全部为\(0\)，包括右下角的一部分全部为\(1\)，于是现在问题转化为求那个分界线是什么。
画一画图，发现每条分界线对应一组割，转化成了最小割模型，然后因为数据范围对\(dinic\)不友好，化成对偶图跑最短路就行了
注意不能只考虑向下和向右的边
代码：

#include 
#include  
#include   
#include   
#include    
#include    
#include    
#include    
#include     
#include     
#include     
#include       
#include       

#define IINF 0x3f3f3f3f3f3f3f3fLL
#define u64 unsigned long long
#define pii pair
#define mii map
#define u32 unsigned int
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define vi vector
#define ll long long
#define mp make_pair
#define pb push_back
#define is insert
#define se second
#define fi first
#define ps push

#define $SHOW(x) cout << #x" = " << x << endl
#define $DEBUG() printf("%d %s\n", __LINE__, __FUNCTION__)

namespace FastIO {
    const int SIZE = (1 << 21) + 1;
    char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55];
    int qr;
    #define gc() (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS++)) : *iS++) // getchar
    inline void flush() { fwrite(obuf, 1, oS - obuf, stdout); oS = obuf; } // print the remaining part
    inline void putc(char x) { *oS++ = x; if (oS == oT) flush(); } // putchar
    inline void wrap() { *oS++ = '\n'; if (oS == oT) flush(); } // wrap
    template 
    inline void gi(I &x) { // input an integer
        I sign = 1;
        for (c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') sign = -1;
        for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15);
        x *= sign;
    }
    inline int gi() { // input an int
        int x, sign = 1;
        for (c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') sign = -1;
        for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15);
        return x * sign;
    }
    inline ll gl() { // input an long long
        ll x, sign = 1;
        for (c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') sign = -1;
        for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15);
        return x * sign;
    }
    template 
    inline void print(I &x) { // print a integer
        if (!x) putc('0');
        while (x) qu[++qr] = x % 10 + '0',  x /= 10;
        while (qr) putc(qu[qr--]);
    }
    template 
    inline void println(I &x) { // print a integer and wrap
        if (!x) putc('0');
        while (x) qu[++qr] = x % 10 + '0',  x /= 10;
        while (qr) putc(qu[qr--]);
        wrap();
    }
}

using namespace std;
// using namespace FastIO;

#define MAXN 250000

struct Edge {
    int next, to, w;
}e[20 * MAXN + 5];

int n, S = MAXN + 1, T = MAXN + 2;
int head[MAXN + 5], eid, d[MAXN + 5];
bool inq[MAXN + 5];

void addEdge(int x, int y, int w) {
    e[++eid].next = head[x];
    e[eid].to = y;
    e[eid].w = w;
    head[x] = eid;
}

void spfa() {
    queue q;
    memset(d, 0x3f, sizeof d);
    d[S] = 0;
    q.push(S);
    inq[S] = 1;
    while(!q.empty()) {
        int u = q.front(); q.pop();
        inq[u] = 0;
        for (int i = head[u]; i; i = e[i].next) {
            int v = e[i].to, w = e[i].w;
            if (d[v] > d[u] + w) {
                d[v] = d[u] + w;
                if (!inq[v]) q.push(v), inq[v] = 1;
            }
        }
    }
}

int main() {
    scanf("%d", &n);
    for (int i = 1, x; i <= n; ++i)
        scanf("%d", &x), addEdge(i, T, x);
    for (int i = 1; i < n; ++i)
        for (int j = 1, x; j <= n; ++j)
            scanf("%d", &x), addEdge(i * n + j, (i - 1) * n + j, x);
    for (int i = 1, x; i <= n; ++i)
        scanf("%d", &x), addEdge(S, (n - 1) * n + i, x);
    for (int i = 1, x; i <= n; ++i) {
        scanf("%d", &x);
        addEdge(S, (i - 1) * n + 1, x);
        for (int j = 1; j < n; ++j)
            scanf("%d", &x), addEdge((i - 1) * n + j, (i - 1) * n + j + 1, x);
        scanf("%d", &x);    
        addEdge(i * n, T, x);
    }
    for (int i = 1, x; i <= n; ++i)
        scanf("%d", &x), addEdge(T, i, x);
    for (int i = 1; i < n; ++i)
        for (int j = 1, x; j <= n; ++j)
            scanf("%d", &x), addEdge((i - 1) * n + j, i * n + j, x);
    for (int i = 1, x; i <= n; ++i)
        scanf("%d", &x), addEdge((n - 1) * n + i, S, x);
    for (int i = 1, x; i <= n; ++i) {
        scanf("%d", &x);
        addEdge((i - 1) * n + 1, S, x);
        for (int j = 1; j < n; ++j)
            scanf("%d", &x), addEdge((i - 1) * n + j + 1, (i - 1) * n + j, x);
        scanf("%d", &x);
        addEdge(T, i * n, x);
    }
    spfa();
    printf("%d\n", d[T]);
    return 0;
}
图论——最短路图论——对偶图最短路图论

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