【DP】 HDOJ 2829 Lawrence

w[i][j]代表i到j中没有炸弹的代价。。。dp[i][j]代表前j个数用了i个炸弹的最小代价 然后显然有dp[i][j] = min(dp[i][j], dp[i-1][k] + w[k+1][j])。。通过打表可知满足四边行不等式。。。然后这个方程可以用四边行不等式优化。。。。

#include <iostream>#include <queue> #include <stack> #include <map> #include <set> #include <bitset> #include <cstdio> #include <algorithm> #include <cstring> #include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 1005#define maxm 1000005#define eps 1e-10#define mod 1000000007#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}// headint dp[maxn][maxn];int w[maxn][maxn];int s[maxn][maxn];int a[maxn];int n, m;void read(void){for(int i = 1; i <= n; i++) scanf("%d", &a[i]);}void debug(void){for(int i = 1; i<= m; i++) {for(int j = 1; j <= n; j++)printf("%lld ", dp[i][j]);printf("\n");}}void work(void){for(int i = 1; i <= n; i++) {int t1 = 0, t2 = 0;for(int j = i; j <= n; j++) {t2 += t1 * a[j];t1 += a[j];w[i][j] = t2;}}for(int i = 1; i <= n; i++) dp[0][i] = w[1][i];for(int i = 1; i <= m; i++) s[i][n+1] = n;for(int i = 1; i <= m; i++)for(int j = n; j >= 1; j--) {dp[i][j] = INF;for(int k = s[i-1][j]; k <= s[i][j+1]; k++)if(dp[i][j] > dp[i-1][k] + w[k+1][j]) {dp[i][j] = dp[i-1][k] + w[k+1][j];s[i][j] = k;}}printf("%d\n", dp[m][n]);}int main(void){while(scanf("%d%d", &n, &m), n != 0 || m != 0) {read();work();}return 0;}