Description
给你一个n*m的矩阵,边界的墙都是H,相邻两个格子会有墙,问你往里面灌水有多少种不同的方案。
Sample Input
3 2 2
1
1
1
1 2
1 1
Sample Output
65
你考虑将边从小到大排序,每次将一条边相邻两个格子和到一个并差集中,并统计答案,对于一个连通块,它们往上的高度一定都是一样的,你可以据此统计答案。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
int _min(int x, int y) {return x < y ? x : y;}
int _max(int x, int y) {return x > y ? x : y;}
int read() {
int s = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
return s * f;
}
struct edge {
int x, y, c;
} e[1100000];
int fa[510000], low[510000];
LL s[510000];
bool cmp(edge a, edge b) {return a.c < b.c;}
int findfa(int x) {
if(fa[x] != x) fa[x] = findfa(fa[x]);
return fa[x];
}
int main() {
int n = read(), m = read(), H = read();
int len = 0;
for(int i = 1; i <= n; i++) {
for(int j = 1; j < m; j++) {
e[++len].x = (i - 1) * m + j, e[len].y = (i - 1) * m + j + 1;
e[len].c = read();
}
}
for(int i = 1; i < n; i++) {
for(int j = 1; j <= m; j++) {
e[++len].x = (i - 1) * m + j, e[len].y = i * m + j;
e[len].c = read();
}
} sort(e + 1, e + len + 1, cmp); LL ans = 0;
for(int i = 1; i <= n * m; i++) low[i] = 0, s[i] = 1, fa[i] = i;
int hh = n * m;
for(int i = 1; i <= len; i++) {
int fx = findfa(e[i].x), fy = findfa(e[i].y);
if(fx != fy) {
fa[fx] = fy;
s[fy] = ((s[fx] + e[i].c - low[fx]) * (s[fy] + e[i].c - low[fy]) % mod) % mod;
low[fy] = e[i].c; hh--;
} if(hh == 1) {(ans += H - e[i].c) %= mod; break;}
} (ans += s[findfa(1)]) %= mod;
printf("%lld\n", ans);
return 0;
}