UVA 12295 Optimal Symmetric Paths（spfa+记忆化）

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题意：
求从左上角到右下角的最短路径数，且要求沿斜线对称
思路：
既然要求对称，所以我们将对称的权值叠加，那么就是求到对角线的最短路径了，通过dp解决方案数

//      whn6325689
//      Mr.Phoebe
//      http://blog.csdn.net/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
#define speed std::ios::sync_with_stdio(false);typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))
#define CPY(x,y) memcpy(x,y,sizeof(x))
#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))
#define ls (idx<<1)
#define rs (idx<<1|1)
#define lson ls,l,mid
#define rson rs,mid+1,r
#define root 1,1,ntemplate<class T>
inline bool read(T &n)
{T x = 0, tmp = 1;char c = getchar();while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();if(c == EOF) return false;if(c == '-') c = getchar(), tmp = -1;while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();n = x*tmp;return true;
}
template <class T>
inline void write(T n)
{if(n < 0){putchar('-');n = -n;}int len = 0,data[20];while(n){data[len++] = n%10;n /= 10;}if(!len) data[len++] = 0;while(len--) putchar(data[len]+48);
}
//-----------------------------------const int MOD=1000000009;int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
ll dp[10010];
int d[10010],vis[10010],minn;
int N,a[111][111];
int q[10010];int init()
{scanf("%d",&N);if(!N)return 0;for(int i=0; i<N; i++)for(int j=0; j<N; j++)scanf("%d",&a[i][j]);for(int i=0; i<N; i++)for(int j=0; j<N-i-1; j ++)a[i][j]+=a[N-1-j][N-1-i];return 1;
}
void spfa()
{int i,j,k,z,next,x,y,xx,yy,front,rear;for(i=0; i<10010; i++)d[i]=1000000000;d[1]=a[0][0];CLR(vis,0);vis[1]=1;front=rear=0;q[rear++]=1;while(front != rear){z = q[front++];if(front > N*N)front=0;vis[z]=0;x=(z-1)/N;y=(z-1)%N;if(x+y == N-1)continue;for(i=0; i<4; i++){xx=x+dx[i];yy=y+dy[i];if(xx>=0 && xx<N && yy>=0 && yy<N){next=xx*N+yy+1;if(d[z]+a[xx][yy] < d[next]){d[next]=d[z]+a[xx][yy];if(!vis[next]){q[rear ++] = next;if(rear > N*N)rear=0;vis[next]=1;}}}}}
}
ll DP(int cur)
{int i,x,y,xx,yy,next;if(dp[cur]!=-1)return dp[cur];x=(cur-1)/N;y=(cur-1)%N;if(x+y == N-1){if(d[cur] == minn)return dp[cur]=1;elsereturn dp[cur]=0;}dp[cur]=0;for(int i=0; i<4; i++){xx=x+dx[i];yy=y+dy[i];if(xx>=0 && xx<N && yy>=0 && yy<N){next=xx*N+yy+1;if(d[cur]+a[xx][yy]==d[next])dp[cur]=(dp[cur]+DP(next))%MOD;}}return dp[cur];
}int main()
{while(init()){spfa();minn=1000000000;for(int i=0; i<N; i++)if(d[i*N+N-i]<minn)minn=d[i*N+N-i];CLR(dp,-1);printf("%lld\n",DP(1));}return 0;
}